GCSE Maths: Factorising x² + bx + c and the difference of two squares (Foundation)

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 5x + 6 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = 6.
   We need two numbers that multiply to 6 (the last number) and add to 5 (the middle number). List the pairs that multiply to 6:
   1 × 6 = 6,   add: 1 + 6 = 7 ✗
   −1 × −6 = 6,   add: −1 + −6 = −7 ✗
   2 × 3 = 6,   add: 2 + 3 = 5 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 3).
   Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 5x + 6 = (x + 2)(x + 3)

Check. Expand the answer: (x + 2)(x + 3) gives x² + 5x + 6 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 7x + 10 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 10.
   We need two numbers that multiply to 10 (the last number) and add to 7 (the middle number). List the pairs that multiply to 10:
   1 × 10 = 10,   add: 1 + 10 = 11 ✗
   −1 × −10 = 10,   add: −1 + −10 = −11 ✗
   2 × 5 = 10,   add: 2 + 5 = 7 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 5).
   Check by expanding: x² + 5x + 2x + 10 = x² + 7x + 10. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 7x + 10 = (x + 2)(x + 5)

Check. Expand the answer: (x + 2)(x + 5) gives x² + 7x + 10 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 8x + 15 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 8 and c = 15.
   We need two numbers that multiply to 15 (the last number) and add to 8 (the middle number). List the pairs that multiply to 15:
   1 × 15 = 15,   add: 1 + 15 = 16 ✗
   −1 × −15 = 15,   add: −1 + −15 = −16 ✗
   3 × 5 = 15,   add: 3 + 5 = 8 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x + 5).
   Check by expanding: x² + 5x + 3x + 15 = x² + 8x + 15. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 8x + 15 = (x + 3)(x + 5)

Check. Expand the answer: (x + 3)(x + 5) gives x² + 8x + 15 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 9x + 20 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 9 and c = 20.
   We need two numbers that multiply to 20 (the last number) and add to 9 (the middle number). List the pairs that multiply to 20:
   1 × 20 = 20,   add: 1 + 20 = 21 ✗
   −1 × −20 = 20,   add: −1 + −20 = −21 ✗
   2 × 10 = 20,   add: 2 + 10 = 12 ✗
   −2 × −10 = 20,   add: −2 + −10 = −12 ✗
   4 × 5 = 20,   add: 4 + 5 = 9 ✓ this pair works
   Put the pair into two brackets: (x + 4)(x + 5).
   Check by expanding: x² + 5x + 4x + 20 = x² + 9x + 20. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 9x + 20 = (x + 4)(x + 5)

Check. Expand the answer: (x + 4)(x + 5) gives x² + 9x + 20 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 10x + 24 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 10 and c = 24.
   We need two numbers that multiply to 24 (the last number) and add to 10 (the middle number). List the pairs that multiply to 24:
   1 × 24 = 24,   add: 1 + 24 = 25 ✗
   −1 × −24 = 24,   add: −1 + −24 = −25 ✗
   2 × 12 = 24,   add: 2 + 12 = 14 ✗
   −2 × −12 = 24,   add: −2 + −12 = −14 ✗
   3 × 8 = 24,   add: 3 + 8 = 11 ✗
   −3 × −8 = 24,   add: −3 + −8 = −11 ✗
   4 × 6 = 24,   add: 4 + 6 = 10 ✓ this pair works
   Put the pair into two brackets: (x + 4)(x + 6).
   Check by expanding: x² + 6x + 4x + 24 = x² + 10x + 24. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 10x + 24 = (x + 4)(x + 6)

Check. Expand the answer: (x + 4)(x + 6) gives x² + 10x + 24 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 6x + 8 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 6 and c = 8.
   We need two numbers that multiply to 8 (the last number) and add to 6 (the middle number). List the pairs that multiply to 8:
   1 × 8 = 8,   add: 1 + 8 = 9 ✗
   −1 × −8 = 8,   add: −1 + −8 = −9 ✗
   2 × 4 = 8,   add: 2 + 4 = 6 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 4).
   Check by expanding: x² + 4x + 2x + 8 = x² + 6x + 8. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 6x + 8 = (x + 2)(x + 4)

Check. Expand the answer: (x + 2)(x + 4) gives x² + 6x + 8 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 7x + 12 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −7 and c = 12.
   We need two numbers that multiply to 12 (the last number) and add to −7 (the middle number). List the pairs that multiply to 12:
   1 × 12 = 12,   add: 1 + 12 = 13 ✗
   −1 × −12 = 12,   add: −1 + −12 = −13 ✗
   2 × 6 = 12,   add: 2 + 6 = 8 ✗
   −2 × −6 = 12,   add: −2 + −6 = −8 ✗
   3 × 4 = 12,   add: 3 + 4 = 7 ✗
   −3 × −4 = 12,   add: −3 + −4 = −7 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x − 4).
   Check by expanding: x² − 4x − 3x + 12 = x² − 7x + 12. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 7x + 12 = (x − 4)(x − 3)

Check. Expand the answer: (x − 4)(x − 3) gives x² − 7x + 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 6x + 8 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −6 and c = 8.
   We need two numbers that multiply to 8 (the last number) and add to −6 (the middle number). List the pairs that multiply to 8:
   1 × 8 = 8,   add: 1 + 8 = 9 ✗
   −1 × −8 = 8,   add: −1 + −8 = −9 ✗
   2 × 4 = 8,   add: 2 + 4 = 6 ✗
   −2 × −4 = 8,   add: −2 + −4 = −6 ✓ this pair works
   Put the pair into two brackets: (x − 2)(x − 4).
   Check by expanding: x² − 4x − 2x + 8 = x² − 6x + 8. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 6x + 8 = (x − 4)(x − 2)

Check. Expand the answer: (x − 4)(x − 2) gives x² − 6x + 8 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 10x + 21 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −10 and c = 21.
   We need two numbers that multiply to 21 (the last number) and add to −10 (the middle number). List the pairs that multiply to 21:
   1 × 21 = 21,   add: 1 + 21 = 22 ✗
   −1 × −21 = 21,   add: −1 + −21 = −22 ✗
   3 × 7 = 21,   add: 3 + 7 = 10 ✗
   −3 × −7 = 21,   add: −3 + −7 = −10 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x − 7).
   Check by expanding: x² − 7x − 3x + 21 = x² − 10x + 21. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 10x + 21 = (x − 7)(x − 3)

Check. Expand the answer: (x − 7)(x − 3) gives x² − 10x + 21 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 8x + 15 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −8 and c = 15.
   We need two numbers that multiply to 15 (the last number) and add to −8 (the middle number). List the pairs that multiply to 15:
   1 × 15 = 15,   add: 1 + 15 = 16 ✗
   −1 × −15 = 15,   add: −1 + −15 = −16 ✗
   3 × 5 = 15,   add: 3 + 5 = 8 ✗
   −3 × −5 = 15,   add: −3 + −5 = −8 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x − 5).
   Check by expanding: x² − 5x − 3x + 15 = x² − 8x + 15. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 8x + 15 = (x − 5)(x − 3)

Check. Expand the answer: (x − 5)(x − 3) gives x² − 8x + 15 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 9x + 14 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −9 and c = 14.
   We need two numbers that multiply to 14 (the last number) and add to −9 (the middle number). List the pairs that multiply to 14:
   1 × 14 = 14,   add: 1 + 14 = 15 ✗
   −1 × −14 = 14,   add: −1 + −14 = −15 ✗
   2 × 7 = 14,   add: 2 + 7 = 9 ✗
   −2 × −7 = 14,   add: −2 + −7 = −9 ✓ this pair works
   Put the pair into two brackets: (x − 2)(x − 7).
   Check by expanding: x² − 7x − 2x + 14 = x² − 9x + 14. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 9x + 14 = (x − 7)(x − 2)

Check. Expand the answer: (x − 7)(x − 2) gives x² − 9x + 14 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 2x − 8 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 2 and c = −8.
   We need two numbers that multiply to −8 (the last number) and add to 2 (the middle number). List the pairs that multiply to −8:
   1 × −8 = −8,   add: 1 + −8 = −7 ✗
   −1 × 8 = −8,   add: −1 + 8 = 7 ✗
   2 × −4 = −8,   add: 2 + −4 = −2 ✗
   −2 × 4 = −8,   add: −2 + 4 = 2 ✓ this pair works
   Put the pair into two brackets: (x − 2)(x + 4).
   Check by expanding: x² + 4x − 2x − 8 = x² + 2x − 8. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 2x − 8 = (x − 2)(x + 4)

Check. Expand the answer: (x − 2)(x + 4) gives x² + 2x − 8 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − x − 12 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −1 and c = −12.
   We need two numbers that multiply to −12 (the last number) and add to −1 (the middle number). List the pairs that multiply to −12:
   1 × −12 = −12,   add: 1 + −12 = −11 ✗
   −1 × 12 = −12,   add: −1 + 12 = 11 ✗
   2 × −6 = −12,   add: 2 + −6 = −4 ✗
   −2 × 6 = −12,   add: −2 + 6 = 4 ✗
   3 × −4 = −12,   add: 3 + −4 = −1 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x − 4).
   Check by expanding: x² − 4x + 3x − 12 = x² − x − 12. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − x − 12 = (x − 4)(x + 3)

Check. Expand the answer: (x − 4)(x + 3) gives x² − x − 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 3x − 18 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = −18.
   We need two numbers that multiply to −18 (the last number) and add to 3 (the middle number). List the pairs that multiply to −18:
   1 × −18 = −18,   add: 1 + −18 = −17 ✗
   −1 × 18 = −18,   add: −1 + 18 = 17 ✗
   2 × −9 = −18,   add: 2 + −9 = −7 ✗
   −2 × 9 = −18,   add: −2 + 9 = 7 ✗
   3 × −6 = −18,   add: 3 + −6 = −3 ✗
   −3 × 6 = −18,   add: −3 + 6 = 3 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x + 6).
   Check by expanding: x² + 6x − 3x − 18 = x² + 3x − 18. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 3x − 18 = (x − 3)(x + 6)

Check. Expand the answer: (x − 3)(x + 6) gives x² + 3x − 18 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 4x − 5 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −4 and c = −5.
   We need two numbers that multiply to −5 (the last number) and add to −4 (the middle number). List the pairs that multiply to −5:
   1 × −5 = −5,   add: 1 + −5 = −4 ✓ this pair works
   Put the pair into two brackets: (x + 1)(x − 5).
   Check by expanding: x² − 5x + 1x − 5 = x² − 4x − 5. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 4x − 5 = (x − 5)(x + 1)

Check. Expand the answer: (x − 5)(x + 1) gives x² − 4x − 5 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 5x − 14 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = −14.
   We need two numbers that multiply to −14 (the last number) and add to 5 (the middle number). List the pairs that multiply to −14:
   1 × −14 = −14,   add: 1 + −14 = −13 ✗
   −1 × 14 = −14,   add: −1 + 14 = 13 ✗
   2 × −7 = −14,   add: 2 + −7 = −5 ✗
   −2 × 7 = −14,   add: −2 + 7 = 5 ✓ this pair works
   Put the pair into two brackets: (x − 2)(x + 7).
   Check by expanding: x² + 7x − 2x − 14 = x² + 5x − 14. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 5x − 14 = (x − 2)(x + 7)

Check. Expand the answer: (x − 2)(x + 7) gives x² + 5x − 14 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 2x − 24 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −2 and c = −24.
   We need two numbers that multiply to −24 (the last number) and add to −2 (the middle number). List the pairs that multiply to −24:
   1 × −24 = −24,   add: 1 + −24 = −23 ✗
   −1 × 24 = −24,   add: −1 + 24 = 23 ✗
   2 × −12 = −24,   add: 2 + −12 = −10 ✗
   −2 × 12 = −24,   add: −2 + 12 = 10 ✗
   3 × −8 = −24,   add: 3 + −8 = −5 ✗
   −3 × 8 = −24,   add: −3 + 8 = 5 ✗
   4 × −6 = −24,   add: 4 + −6 = −2 ✓ this pair works
   Put the pair into two brackets: (x + 4)(x − 6).
   Check by expanding: x² − 6x + 4x − 24 = x² − 2x − 24. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 2x − 24 = (x − 6)(x + 4)

Check. Expand the answer: (x − 6)(x + 4) gives x² − 2x − 24 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 9x + 14 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 9 and c = 14.
   We need two numbers that multiply to 14 (the last number) and add to 9 (the middle number). List the pairs that multiply to 14:
   1 × 14 = 14,   add: 1 + 14 = 15 ✗
   −1 × −14 = 14,   add: −1 + −14 = −15 ✗
   2 × 7 = 14,   add: 2 + 7 = 9 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 7).
   Check by expanding: x² + 7x + 2x + 14 = x² + 9x + 14. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 9x + 14 = (x + 2)(x + 7)

Check. Expand the answer: (x + 2)(x + 7) gives x² + 9x + 14 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 8x + 15 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −8 and c = 15.
   We need two numbers that multiply to 15 (the last number) and add to −8 (the middle number). List the pairs that multiply to 15:
   1 × 15 = 15,   add: 1 + 15 = 16 ✗
   −1 × −15 = 15,   add: −1 + −15 = −16 ✗
   3 × 5 = 15,   add: 3 + 5 = 8 ✗
   −3 × −5 = 15,   add: −3 + −5 = −8 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x − 5).
   Check by expanding: x² − 5x − 3x + 15 = x² − 8x + 15. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 8x + 15 = (x − 5)(x − 3)

Check. Expand the answer: (x − 5)(x − 3) gives x² − 8x + 15 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + x − 12 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = −12.
   We need two numbers that multiply to −12 (the last number) and add to 1 (the middle number). List the pairs that multiply to −12:
   1 × −12 = −12,   add: 1 + −12 = −11 ✗
   −1 × 12 = −12,   add: −1 + 12 = 11 ✗
   2 × −6 = −12,   add: 2 + −6 = −4 ✗
   −2 × 6 = −12,   add: −2 + 6 = 4 ✗
   3 × −4 = −12,   add: 3 + −4 = −1 ✗
   −3 × 4 = −12,   add: −3 + 4 = 1 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x + 4).
   Check by expanding: x² + 4x − 3x − 12 = x² + x − 12. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + x − 12 = (x − 3)(x + 4)

Check. Expand the answer: (x − 3)(x + 4) gives x² + x − 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of y² + 8y + 12 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form y² + by + c, with b = 8 and c = 12.
   We need two numbers that multiply to 12 (the last number) and add to 8 (the middle number). List the pairs that multiply to 12:
   1 × 12 = 12,   add: 1 + 12 = 13 ✗
   −1 × −12 = 12,   add: −1 + −12 = −13 ✗
   2 × 6 = 12,   add: 2 + 6 = 8 ✓ this pair works
   Put the pair into two brackets: (y + 2)(y + 6).
   Check by expanding: y² + 6y + 2y + 12 = y² + 8y + 12. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   y² + 8y + 12 = (y + 2)(y + 6)

Check. Expand the answer: (y + 2)(y + 6) gives y² + 8y + 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of a² − a − 20 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form a² + ba + c, with b = −1 and c = −20.
   We need two numbers that multiply to −20 (the last number) and add to −1 (the middle number). List the pairs that multiply to −20:
   1 × −20 = −20,   add: 1 + −20 = −19 ✗
   −1 × 20 = −20,   add: −1 + 20 = 19 ✗
   2 × −10 = −20,   add: 2 + −10 = −8 ✗
   −2 × 10 = −20,   add: −2 + 10 = 8 ✗
   4 × −5 = −20,   add: 4 + −5 = −1 ✓ this pair works
   Put the pair into two brackets: (a + 4)(a − 5).
   Check by expanding: a² − 5a + 4a − 20 = a² − a − 20. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   a² − a − 20 = (a − 5)(a + 4)

Check. Expand the answer: (a − 5)(a + 4) gives a² − a − 20 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of p² − 10p + 16 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form p² + bp + c, with b = −10 and c = 16.
   We need two numbers that multiply to 16 (the last number) and add to −10 (the middle number). List the pairs that multiply to 16:
   1 × 16 = 16,   add: 1 + 16 = 17 ✗
   −1 × −16 = 16,   add: −1 + −16 = −17 ✗
   2 × 8 = 16,   add: 2 + 8 = 10 ✗
   −2 × −8 = 16,   add: −2 + −8 = −10 ✓ this pair works
   Put the pair into two brackets: (p − 2)(p − 8).
   Check by expanding: p² − 8p − 2p + 16 = p² − 10p + 16. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   p² − 10p + 16 = (p − 8)(p − 2)

Check. Expand the answer: (p − 8)(p − 2) gives p² − 10p + 16 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of m² + 4m − 21 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form m² + bm + c, with b = 4 and c = −21.
   We need two numbers that multiply to −21 (the last number) and add to 4 (the middle number). List the pairs that multiply to −21:
   1 × −21 = −21,   add: 1 + −21 = −20 ✗
   −1 × 21 = −21,   add: −1 + 21 = 20 ✗
   3 × −7 = −21,   add: 3 + −7 = −4 ✗
   −3 × 7 = −21,   add: −3 + 7 = 4 ✓ this pair works
   Put the pair into two brackets: (m − 3)(m + 7).
   Check by expanding: m² + 7m − 3m − 21 = m² + 4m − 21. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   m² + 4m − 21 = (m − 3)(m + 7)

Check. Expand the answer: (m − 3)(m + 7) gives m² + 4m − 21 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 6x + 9 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 6 and c = 9.
   We need two numbers that multiply to 9 (the last number) and add to 6 (the middle number). List the pairs that multiply to 9:
   1 × 9 = 9,   add: 1 + 9 = 10 ✗
   −1 × −9 = 9,   add: −1 + −9 = −10 ✗
   3 × 3 = 9,   add: 3 + 3 = 6 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x + 3).
   Check by expanding: x² + 3x + 3x + 9 = x² + 6x + 9. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 6x + 9 = (x + 3)²

Check. Expand the answer: (x + 3)² gives x² + 6x + 9 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 10x + 25 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −10 and c = 25.
   We need two numbers that multiply to 25 (the last number) and add to −10 (the middle number). List the pairs that multiply to 25:
   1 × 25 = 25,   add: 1 + 25 = 26 ✗
   −1 × −25 = 25,   add: −1 + −25 = −26 ✗
   5 × 5 = 25,   add: 5 + 5 = 10 ✗
   −5 × −5 = 25,   add: −5 + −5 = −10 ✓ this pair works
   Put the pair into two brackets: (x − 5)(x − 5).
   Check by expanding: x² − 5x − 5x + 25 = x² − 10x + 25. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − 10x + 25 = (x − 5)²

Check. Expand the answer: (x − 5)² gives x² − 10x + 25 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 14x + 49 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 14 and c = 49.
   We need two numbers that multiply to 49 (the last number) and add to 14 (the middle number). List the pairs that multiply to 49:
   1 × 49 = 49,   add: 1 + 49 = 50 ✗
   −1 × −49 = 49,   add: −1 + −49 = −50 ✗
   7 × 7 = 49,   add: 7 + 7 = 14 ✓ this pair works
   Put the pair into two brackets: (x + 7)(x + 7).
   Check by expanding: x² + 7x + 7x + 49 = x² + 14x + 49. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 14x + 49 = (x + 7)²

Check. Expand the answer: (x + 7)² gives x² + 14x + 49 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of y² − 4y + 4 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form y² + by + c, with b = −4 and c = 4.
   We need two numbers that multiply to 4 (the last number) and add to −4 (the middle number). List the pairs that multiply to 4:
   1 × 4 = 4,   add: 1 + 4 = 5 ✗
   −1 × −4 = 4,   add: −1 + −4 = −5 ✗
   2 × 2 = 4,   add: 2 + 2 = 4 ✗
   −2 × −2 = 4,   add: −2 + −2 = −4 ✓ this pair works
   Put the pair into two brackets: (y − 2)(y − 2).
   Check by expanding: y² − 2y − 2y + 4 = y² − 4y + 4. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   y² − 4y + 4 = (y − 2)²

Check. Expand the answer: (y − 2)² gives y² − 4y + 4 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 8x + 15 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 8 and c = 15.
   We need two numbers that multiply to 15 (the last number) and add to 8 (the middle number). List the pairs that multiply to 15:
   1 × 15 = 15,   add: 1 + 15 = 16 ✗
   −1 × −15 = 15,   add: −1 + −15 = −16 ✗
   3 × 5 = 15,   add: 3 + 5 = 8 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x + 5).
   Check by expanding: x² + 5x + 3x + 15 = x² + 8x + 15. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 8x + 15 = (x + 3)(x + 5)

Check. Expand the answer: (x + 3)(x + 5) gives x² + 8x + 15 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A term is one part of an expression, together with the sign in front of it (for example 3x or −5). Like terms have exactly the same letters with exactly the same powers (3x and 5x are like terms, 3x and 5x² are not). The coefficient is the number in front of the letters.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

1. Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
   +x, +3, +x, +5, +x, +3, +x, +5
2. Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
   x terms: +x, +x, +x, +x
      add the coefficients: 1 + 1 + 1 + 1 = 4  so they give 4x
   number terms: +3, +5, +3, +5
      add the coefficients: 3 + 5 + 3 + 5 = 16  so they give 16
3. Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
   4x + 16

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 28 and the answer gives 28. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Method. List the pairs of numbers that multiply to 12, and see which pair adds to 7.

1. List the factor pairs of 12. We need two numbers that multiply to 12.
   1 × 12,   2 × 6,   3 × 4
2. Find their sums. We need the sum to be 7.
   1 + 12 = 13,   2 + 6 = 8,   3 + 4 = 7 ✓
3. Write the answer.
   The numbers are 3 and 4.

Check. 3 × 4 = 12 and 3 + 4 = 7. ✓

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² + 7x + 12 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
   We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
   1 × 12 = 12,   add: 1 + 12 = 13 ✗
   −1 × −12 = 12,   add: −1 + −12 = −13 ✗
   2 × 6 = 12,   add: 2 + 6 = 8 ✗
   −2 × −6 = 12,   add: −2 + −6 = −8 ✗
   3 × 4 = 12,   add: 3 + 4 = 7 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x + 4).
   Check by expanding: x² + 4x + 3x + 12 = x² + 7x + 12. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² + 7x + 12 = (x + 3)(x + 4)

Check. Expand the answer: (x + 3)(x + 4) gives x² + 7x + 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 9:   x² is a square and 9 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   9 = (3)², so b = 3
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (x − 3)(x + 3)

Check. Expand to check. (x − 3)(x + 3): x × x = x²; x × 3 = 3x and (−3) × x = −3x (these two cancel); (−3) × 3 = −9. Total: x² − 9. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 25:   x² is a square and 25 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   25 = (5)², so b = 5
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (x − 5)(x + 5)

Check. Expand to check. (x − 5)(x + 5): x × x = x²; x × 5 = 5x and (−5) × x = −5x (these two cancel); (−5) × 5 = −25. Total: x² − 25. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   y² − 49:   y² is a square and 49 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   y² = (y)², so a = y
   49 = (7)², so b = 7
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (y − 7)(y + 7)

Check. Expand to check. (y − 7)(y + 7): y × y = y²; y × 7 = 7y and (−7) × y = −7y (these two cancel); (−7) × 7 = −49. Total: y² − 49. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   a² − 100:   a² is a square and 100 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   a² = (a)², so a = a
   100 = (10)², so b = 10
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (a − 10)(a + 10)

Check. Expand to check. (a − 10)(a + 10): a × a = a²; a × 10 = 10a and (−10) × a = −10a (these two cancel); (−10) × 10 = −100. Total: a² − 100. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   p² − 1:   p² is a square and 1 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   p² = (p)², so a = p
   1 = (1)², so b = 1
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (p − 1)(p + 1)

Check. Expand to check. (p − 1)(p + 1): p × p = p²; p × 1 = p and (−1) × p = −p (these two cancel); (−1) × 1 = −1. Total: p² − 1. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 64:   x² is a square and 64 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   64 = (8)², so b = 8
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (x − 8)(x + 8)

Check. Expand to check. (x − 8)(x + 8): x × x = x²; x × 8 = 8x and (−8) × x = −8x (these two cancel); (−8) × 8 = −64. Total: x² − 64. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   4x² − 9:   4x² is a square and 9 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   4x² = (2x)², so a = 2x
   9 = (3)², so b = 3
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (2x − 3)(2x + 3)

Check. Expand to check. (2x − 3)(2x + 3): 2x × 2x = 4x²; 2x × 3 = 6x and (−3) × 2x = −6x (these two cancel); (−3) × 3 = −9. Total: 4x² − 9. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   9y² − 1:   9y² is a square and 1 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   9y² = (3y)², so a = 3y
   1 = (1)², so b = 1
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (3y − 1)(3y + 1)

Check. Expand to check. (3y − 1)(3y + 1): 3y × 3y = 9y²; 3y × 1 = 3y and (−1) × 3y = −3y (these two cancel); (−1) × 1 = −1. Total: 9y² − 1. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   16a² − 25:   16a² is a square and 25 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   16a² = (4a)², so a = 4a
   25 = (5)², so b = 5
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (4a − 5)(4a + 5)

Check. Expand to check. (4a − 5)(4a + 5): 4a × 4a = 16a²; 4a × 5 = 20a and (−5) × 4a = −20a (these two cancel); (−5) × 5 = −25. Total: 16a² − 25. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   100x² − 81:   100x² is a square and 81 is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   100x² = (10x)², so a = 10x
   81 = (9)², so b = 9
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (10x − 9)(10x + 9)

Check. Expand to check. (10x − 9)(10x + 9): 10x × 10x = 100x²; 10x × 9 = 90x and (−9) × 10x = −90x (these two cancel); (−9) × 9 = −81. Total: 100x² − 81. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Method. Use the difference of two squares, a² − b² = (a − b)(a + b), to avoid awkward squaring.

1. Spot the pattern. Two square numbers with a minus between them: this is a difference of two squares, a² − b². Using the rule saves us from squaring big or awkward numbers.
   51² − 49², with a = 51 and b = 49
2. Use the rule a² − b² = (a − b)(a + b). The first bracket is the difference of the two numbers and the second is their sum.
   (a − b) = 51 − 49 = 2
   (a + b) = 51 + 49 = 100
3. Multiply the two answers. These are easy numbers, which is why the rule helps.
   2 × 100 = 200

Check. Calculator check: 51² = 2601 and 49² = 2401, and 2601 − 2401 = 200. ✓

Watch out. Subtract and add the two numbers (not their squares): (a − b) and (a + b).

Method. Use the difference of two squares, a² − b² = (a − b)(a + b), to avoid awkward squaring.

1. Spot the pattern. Two square numbers with a minus between them: this is a difference of two squares, a² − b². Using the rule saves us from squaring big or awkward numbers.
   99² − 1², with a = 99 and b = 1
2. Use the rule a² − b² = (a − b)(a + b). The first bracket is the difference of the two numbers and the second is their sum.
   (a − b) = 99 − 1 = 98
   (a + b) = 99 + 1 = 100
3. Multiply the two answers. These are easy numbers, which is why the rule helps.
   98 × 100 = 9800

Check. Calculator check: 99² = 9801 and 1² = 1, and 9801 − 1 = 9800. ✓

Watch out. Subtract and add the two numbers (not their squares): (a − b) and (a + b).

Method. Use the difference of two squares, a² − b² = (a − b)(a + b), to avoid awkward squaring.

1. Spot the pattern. Two square numbers with a minus between them: this is a difference of two squares, a² − b². Using the rule saves us from squaring big or awkward numbers.
   7.5² − 2.5², with a = 7.5 and b = 2.5
2. Use the rule a² − b² = (a − b)(a + b). The first bracket is the difference of the two numbers and the second is their sum.
   (a − b) = 7.5 − 2.5 = 5
   (a + b) = 7.5 + 2.5 = 10
3. Multiply the two answers. These are easy numbers, which is why the rule helps.
   5 × 10 = 50

Check. Calculator check: 7.5² = 56.25 and 2.5² = 6.25, and 56.25 − 6.25 = 50. ✓

Watch out. Subtract and add the two numbers (not their squares): (a − b) and (a + b).

Method. Use the difference of two squares, a² − b² = (a − b)(a + b), to avoid awkward squaring.

1. Spot the pattern. Two square numbers with a minus between them: this is a difference of two squares, a² − b². Using the rule saves us from squaring big or awkward numbers.
   1001² − 999², with a = 1001 and b = 999
2. Use the rule a² − b² = (a − b)(a + b). The first bracket is the difference of the two numbers and the second is their sum.
   (a − b) = 1001 − 999 = 2
   (a + b) = 1001 + 999 = 2000
3. Multiply the two answers. These are easy numbers, which is why the rule helps.
   2 × 2000 = 4000

Check. Calculator check: 1001² = 1002001 and 999² = 998001, and 1002001 − 998001 = 4000. ✓

Watch out. Subtract and add the two numbers (not their squares): (a − b) and (a + b).

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 2x², −8
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 2: 1, 2
   Factors of 8: 1, 2, 4, 8
   The biggest number that is in every list is 2.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   x: it is missing from at least one term (a missing letter has power 0), so it is not common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 2
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   2x² ÷ 2 = x²
   −8 ÷ 2 = −4
   So we can write 2(x² − 4)
6. Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
   The bracket is (x² − 4) = x² − 4.
7. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 4:   x² is a square and 4 is a square, with a minus between them. ✓
8. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
9. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   4 = (2)², so b = 2
10. Put a and b into the rule. One bracket has a minus, the other has a plus.
    (x − 2)(x + 2)
11. Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
    2x² − 8 = 2(x − 2)(x + 2)

Check. Expand to check. (x − 2)(x + 2): x × x = x²; x × 2 = 2x and (−2) × x = −2x (these two cancel); (−2) × 2 = −4. Total: x² − 4. ✓ Then multiply by 2: 2x² − 8. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 3x², −75
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 3: 1, 3
   Factors of 75: 1, 3, 5, 15, 25, 75
   The biggest number that is in every list is 3.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   x: it is missing from at least one term (a missing letter has power 0), so it is not common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 3
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   3x² ÷ 3 = x²
   −75 ÷ 3 = −25
   So we can write 3(x² − 25)
6. Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
   The bracket is (x² − 25) = x² − 25.
7. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 25:   x² is a square and 25 is a square, with a minus between them. ✓
8. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
9. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   25 = (5)², so b = 5
10. Put a and b into the rule. One bracket has a minus, the other has a plus.
    (x − 5)(x + 5)
11. Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
    3x² − 75 = 3(x − 5)(x + 5)

Check. Expand to check. (x − 5)(x + 5): x × x = x²; x × 5 = 5x and (−5) × x = −5x (these two cancel); (−5) × 5 = −25. Total: x² − 25. ✓ Then multiply by 3: 3x² − 75. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 5y², −20
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 5: 1, 5
   Factors of 20: 1, 2, 4, 5, 10, 20
   The biggest number that is in every list is 5.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   y: it is missing from at least one term (a missing letter has power 0), so it is not common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 5
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   5y² ÷ 5 = y²
   −20 ÷ 5 = −4
   So we can write 5(y² − 4)
6. Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
   The bracket is (y² − 4) = y² − 4.
7. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   y² − 4:   y² is a square and 4 is a square, with a minus between them. ✓
8. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
9. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   y² = (y)², so a = y
   4 = (2)², so b = 2
10. Put a and b into the rule. One bracket has a minus, the other has a plus.
    (y − 2)(y + 2)
11. Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
    5y² − 20 = 5(y − 2)(y + 2)

Check. Expand to check. (y − 2)(y + 2): y × y = y²; y × 2 = 2y and (−2) × y = −2y (these two cancel); (−2) × 2 = −4. Total: y² − 4. ✓ Then multiply by 5: 5y² − 20. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 4a², −100
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 4: 1, 2, 4
   Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100
   The biggest number that is in every list is 4.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   a: it is missing from at least one term (a missing letter has power 0), so it is not common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 4
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   4a² ÷ 4 = a²
   −100 ÷ 4 = −25
   So we can write 4(a² − 25)
6. Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
   The bracket is (a² − 25) = a² − 25.
7. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   a² − 25:   a² is a square and 25 is a square, with a minus between them. ✓
8. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
9. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   a² = (a)², so a = a
   25 = (5)², so b = 5
10. Put a and b into the rule. One bracket has a minus, the other has a plus.
    (a − 5)(a + 5)
11. Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
    4a² − 100 = 4(a − 5)(a + 5)

Check. Expand to check. (a − 5)(a + 5): a × a = a²; a × 5 = 5a and (−5) × a = −5a (these two cancel); (−5) × 5 = −25. Total: a² − 25. ✓ Then multiply by 4: 4a² − 100. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 7x², −7
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 7: 1, 7
   Factors of 7: 1, 7
   The biggest number that is in every list is 7.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   x: it is missing from at least one term (a missing letter has power 0), so it is not common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 7
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   7x² ÷ 7 = x²
   −7 ÷ 7 = −1
   So we can write 7(x² − 1)
6. Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
   The bracket is (x² − 1) = x² − 1.
7. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 1:   x² is a square and 1 is a square, with a minus between them. ✓
8. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
9. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   1 = (1)², so b = 1
10. Put a and b into the rule. One bracket has a minus, the other has a plus.
    (x − 1)(x + 1)
11. Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
    7x² − 7 = 7(x − 1)(x + 1)

Check. Expand to check. (x − 1)(x + 1): x × x = x²; x × 1 = x and (−1) × x = −x (these two cancel); (−1) × 1 = −1. Total: x² − 1. ✓ Then multiply by 7: 7x² − 7. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − y²:   x² is a square and y² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   y² = (y)², so b = y
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (x − y)(x + y)

Check. Expand to check. (x − y)(x + y): x × x = x²; x × y = xy and (−y) × x = −xy (these two cancel); (−y) × y = −y². Total: x² − y². ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   9a² − 4b²:   9a² is a square and 4b² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   9a² = (3a)², so a = 3a
   4b² = (2b)², so b = 2b
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (3a − 2b)(3a + 2b)

Check. Expand to check. (3a − 2b)(3a + 2b): 3a × 3a = 9a²; 3a × 2b = 6ab and (−2b) × 3a = −6ab (these two cancel); (−2b) × 2b = −4b². Total: 9a² − 4b². ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   25p² − 16q²:   25p² is a square and 16q² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   25p² = (5p)², so a = 5p
   16q² = (4q)², so b = 4q
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (5p − 4q)(5p + 4q)

Check. Expand to check. (5p − 4q)(5p + 4q): 5p × 5p = 25p²; 5p × 4q = 20pq and (−4q) × 5p = −20pq (these two cancel); (−4q) × 4q = −16q². Total: 25p² − 16q². ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   49m² − n²:   49m² is a square and n² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   49m² = (7m)², so a = 7m
   n² = (n)², so b = n
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (7m − n)(7m + n)

Check. Expand to check. (7m − n)(7m + n): 7m × 7m = 49m²; 7m × n = 7mn and (−n) × 7m = −7mn (these two cancel); (−n) × n = −n². Total: 49m² − n². ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   25 − x²:   25 is a square and x² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   25 = (5)², so a = 5
   x² = (x)², so b = x
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (5 − x)(5 + x)

Check. Expand to check. (5 − x)(5 + x): 5 × 5 = 25; 5 × x = 5x and (−x) × 5 = −5x (these two cancel); (−x) × x = −x². Total: −x² + 25. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   49 − 4y²:   49 is a square and 4y² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   49 = (7)², so a = 7
   4y² = (2y)², so b = 2y
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (7 − 2y)(7 + 2y)

Check. Expand to check. (7 − 2y)(7 + 2y): 7 × 7 = 49; 7 × 2y = 14y and (−2y) × 7 = −14y (these two cancel); (−2y) × 2y = −4y². Total: −4y² + 49. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   1 − 9a²:   1 is a square and 9a² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   1 = (1)², so a = 1
   9a² = (3a)², so b = 3a
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (1 − 3a)(1 + 3a)

Check. Expand to check. (1 − 3a)(1 + 3a): 1 × 1 = 1; 1 × 3a = 3a and (−3a) × 1 = −3a (these two cancel); (−3a) × 3a = −9a². Total: −9a² + 1. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Method. Area left = area of the big square − area of the small square.

1. Big square. Side 10, so area = 10 × 10.
   100
2. Small square. Side x, so area = x × x.
   x²
3. Subtract.
   100 − x²

Check. If x = 4: 100 − 16 = 84, and 10 × 10 − 4 × 4 = 84. ✓

Key words. A square number is a number multiplied by itself (49 = 7 × 7). A perfect square term such as 9x² = (3x)² is also something multiplied by itself.

Method. Difference of two squares: a² − b² = (a − b)(a + b). Find a and b by square rooting both terms.

1. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   100 − x²:   100 is a square and x² is a square, with a minus between them. ✓
2. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
3. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   100 = (10)², so a = 10
   x² = (x)², so b = x
4. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (10 − x)(10 + x)

Check. Expand to check. (10 − x)(10 + x): 10 × 10 = 100; 10 × x = 10x and (−x) × 10 = −10x (these two cancel); (−x) × x = −x². Total: −x² + 100. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Always look for a common factor first. Take out the highest common factor, then see whether the bracket can be factorised further.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: x², 9x
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 1: 1
   Factors of 9: 1, 3, 9
   The biggest number that is in every list is 1.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   x: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so x is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = x
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   x² ÷ x = x
   9x ÷ x = 9
   So we can write x(x + 9)
   The bracket has no common factor left and cannot be factorised again, so this is the answer.

Check. Expand: x(x + 9) = x² + 9x. ✓

Watch out. Take out the highest common factor. If you take out only 2 from 6x + 12, the bracket (3x + 6) still has a common factor, so it is not fully factorised. And if a term is the common factor itself, the bracket gets a 1, not a blank: 3x + 3 = 3(x + 1).

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − x − 20 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −1 and c = −20.
   We need two numbers that multiply to −20 (the last number) and add to −1 (the middle number). List the pairs that multiply to −20:
   1 × −20 = −20,   add: 1 + −20 = −19 ✗
   −1 × 20 = −20,   add: −1 + 20 = 19 ✗
   2 × −10 = −20,   add: 2 + −10 = −8 ✗
   −2 × 10 = −20,   add: −2 + 10 = 8 ✗
   4 × −5 = −20,   add: 4 + −5 = −1 ✓ this pair works
   Put the pair into two brackets: (x + 4)(x − 5).
   Check by expanding: x² − 5x + 4x − 20 = x² − x − 20. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   x² − x − 20 = (x − 5)(x + 4)

Check. Expand the answer: (x − 5)(x + 4) gives x² − x − 20 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of x² − 36 have no common factor other than 1, so there is nothing to take out.
2. Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
   x² − 36:   x² is a square and 36 is a square, with a minus between them. ✓
3. Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
   a² − b² = (a − b)(a + b)
4. Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
   x² = (x)², so a = x
   36 = (6)², so b = 6
5. Put a and b into the rule. One bracket has a minus, the other has a plus.
   (x − 6)(x + 6)

Check. Expand to check. (x − 6)(x + 6): x × x = x²; x × 6 = 6x and (−6) × x = −6x (these two cancel); (−6) × 6 = −36. Total: x² − 36. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Always look for a common factor first. Take out the highest common factor, then see whether the bracket can be factorised further.

1. Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
   The terms are: 3x², 6x
2. Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
   Factors of 3: 1, 3
   Factors of 6: 1, 2, 3, 6
   The biggest number that is in every list is 3.
3. Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
   x: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so x is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 3x
5. Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
   3x² ÷ 3x = x
   6x ÷ 3x = 2
   So we can write 3x(x + 2)
   The bracket has no common factor left and cannot be factorised again, so this is the answer.

Check. Expand: 3x(x + 2) = 3x² + 6x. ✓

Watch out. Take out the highest common factor. If you take out only 2 from 6x + 12, the bracket (3x + 6) still has a common factor, so it is not fully factorised. And if a term is the common factor itself, the bracket gets a 1, not a blank: 3x + 3 = 3(x + 1).

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

1. Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
   The terms of y² − 2y − 24 have no common factor other than 1, so there is nothing to take out.
2. Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
   Spot the pattern. This is a quadratic of the form y² + by + c, with b = −2 and c = −24.
   We need two numbers that multiply to −24 (the last number) and add to −2 (the middle number). List the pairs that multiply to −24:
   1 × −24 = −24,   add: 1 + −24 = −23 ✗
   −1 × 24 = −24,   add: −1 + 24 = 23 ✗
   2 × −12 = −24,   add: 2 + −12 = −10 ✗
   −2 × 12 = −24,   add: −2 + 12 = 10 ✗
   3 × −8 = −24,   add: 3 + −8 = −5 ✗
   −3 × 8 = −24,   add: −3 + 8 = 5 ✗
   4 × −6 = −24,   add: 4 + −6 = −2 ✓ this pair works
   Put the pair into two brackets: (y + 4)(y − 6).
   Check by expanding: y² − 6y + 4y − 24 = y² − 2y − 24. ✓
3. Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
   y² − 2y − 24 = (y − 6)(y + 4)