GCSE Maths: Factorising: common factors (Higher)

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 4x, 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 4: 1, 2, 4
   Factors of 8: 1, 2, 4, 8
   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).
   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 = 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.
   4x ÷ 4 = x
   8 ÷ 4 = 2
   So we can write 4(x + 2)

Check. Expand the answer to see if you get the original back: 4 × x = 4x; 4 × 2 = 8. Together: 4x + 8. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 6y, −9
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 6: 1, 2, 3, 6
   Factors of 9: 1, 3, 9
   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).
   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 = 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.
   6y ÷ 3 = 2y
   −9 ÷ 3 = −3
   So we can write 3(2y − 3)

Check. Expand the answer to see if you get the original back: 3 × 2y = 6y; 3 × (−3) = −9. Together: 6y − 9. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 15a, 10
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 15: 1, 3, 5, 15
   Factors of 10: 1, 2, 5, 10
   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).
   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 = 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.
   15a ÷ 5 = 3a
   10 ÷ 5 = 2
   So we can write 5(3a + 2)

Check. Expand the answer to see if you get the original back: 5 × 3a = 15a; 5 × 2 = 10. Together: 15a + 10. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: −8b, 12
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 8: 1, 2, 4, 8
   Factors of 12: 1, 2, 3, 4, 6, 12
   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).
   b: 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.
   −8b ÷ 4 = −2b
   12 ÷ 4 = 3
   So we can write 4(−2b + 3)

Check. Expand the answer to see if you get the original back: 4 × (−2b) = −8b; 4 × 3 = 12. Together: −8b + 12. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 14m, 21
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 14: 1, 2, 7, 14
   Factors of 21: 1, 3, 7, 21
   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).
   m: 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.
   14m ÷ 7 = 2m
   21 ÷ 7 = 3
   So we can write 7(2m + 3)

Check. Expand the answer to see if you get the original back: 7 × 2m = 14m; 7 × 3 = 21. Together: 14m + 21. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 9p, −27
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 9: 1, 3, 9
   Factors of 27: 1, 3, 9, 27
   The biggest number that is in every list is 9.
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).
   p: 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 = 9
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.
   9p ÷ 9 = p
   −27 ÷ 9 = −3
   So we can write 9(p − 3)

Check. Expand the answer to see if you get the original back: 9 × p = 9p; 9 × (−3) = −27. Together: 9p − 27. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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, 3x
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 3: 1, 3
   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
   3x ÷ x = 3
   So we can write x(x + 3)

Check. Expand the answer to see if you get the original back: x × x = x²; x × 3 = 3x. Together: x² + 3x. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: y², −4y
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 4: 1, 2, 4
   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).
   y: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so y is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = y
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.
   y² ÷ y = y
   −4y ÷ y = −4
   So we can write y(y − 4)

Check. Expand the answer to see if you get the original back: y × y = y²; y × (−4) = −4y. Together: y² − 4y. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: a², a
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 1: 1
   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).
   a: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so a is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = a
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.
   a² ÷ a = a
   a ÷ a = 1
   So we can write a(a + 1)

Check. Expand the answer to see if you get the original back: a × a = a²; a × 1 = a. Together: a² + a. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 5m², −m
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 1: 1
   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).
   m: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so m is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = m
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.
   5m² ÷ m = 5m
   −m ÷ m = −1
   So we can write m(5m − 1)

Check. Expand the answer to see if you get the original back: m × 5m = 5m²; m × (−1) = −m. Together: 5m² − m. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 6x², 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 6: 1, 2, 3, 6
   Factors of 9: 1, 3, 9
   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.
   6x² ÷ 3x = 2x
   9x ÷ 3x = 3
   So we can write 3x(2x + 3)

Check. Expand the answer to see if you get the original back: 3x × 2x = 6x²; 3x × 3 = 9x. Together: 6x² + 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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 10a², −15a
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 10: 1, 2, 5, 10
   Factors of 15: 1, 3, 5, 15
   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).
   a: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so a is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 5a
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.
   10a² ÷ 5a = 2a
   −15a ÷ 5a = −3
   So we can write 5a(2a − 3)

Check. Expand the answer to see if you get the original back: 5a × 2a = 10a²; 5a × (−3) = −15a. Together: 10a² − 15a. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 4y², 12y
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 12: 1, 2, 3, 4, 6, 12
   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).
   y: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so y is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 4y
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.
   4y² ÷ 4y = y
   12y ÷ 4y = 3
   So we can write 4y(y + 3)

Check. Expand the answer to see if you get the original back: 4y × y = 4y²; 4y × 3 = 12y. Together: 4y² + 12y. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 8m², −2m
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 8: 1, 2, 4, 8
   Factors of 2: 1, 2
   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).
   m: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so m is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 2m
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.
   8m² ÷ 2m = 4m
   −2m ÷ 2m = −1
   So we can write 2m(4m − 1)

Check. Expand the answer to see if you get the original back: 2m × 4m = 8m²; 2m × (−1) = −2m. Together: 8m² − 2m. ✓

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).

Method. Divide each term of the original expression by the number outside.

1. Find what is left from the 20. 5 × ? = 20.
   20 ÷ 5 = 4
2. Check the first term. 5 × x = 5x, which matches.

Check. 5(x + 4) = 5x + 20. ✓

Method. Divide each term of the original expression by the number outside.

1. Find what goes with the first term. 4 × ? = 8y.
   8y ÷ 4 = 2y
2. Check the second term. 4 × 3 = 12, which matches.

Check. 4(2y − 3) = 8y − 12. ✓

Method. Divide each term by x.

1. Find what goes with the first term. x × ? = x².
   x² ÷ x = x
2. Check the second term. x × 6 = 6x, which matches.

Check. x(x + 6) = x² + 6x. ✓

Method. Divide each term by 3a.

1. Find what goes with the second term. 3a × ? = 12a.
   12a ÷ 3a = 4
2. Check the first term. 3a × a = 3a², which matches.

Check. 3a(a + 4) = 3a² + 12a. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 6ab, 9a
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 6: 1, 2, 3, 6
   Factors of 9: 1, 3, 9
   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).
   a: the powers in the terms are 1, 1. It is in every term and the lowest power is 1, so a is common.
   b: 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 = 3a
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.
   6ab ÷ 3a = 2b
   9a ÷ 3a = 3
   So we can write 3a(2b + 3)

Check. Expand the answer to see if you get the original back: 3a × 2b = 6ab; 3a × 3 = 9a. Together: 6ab + 9a. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 4pq, −6q
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 6: 1, 2, 3, 6
   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).
   p: it is missing from at least one term (a missing letter has power 0), so it is not common.
   q: the powers in the terms are 1, 1. It is in every term and the lowest power is 1, so q is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 2q
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.
   4pq ÷ 2q = 2p
   −6q ÷ 2q = −3
   So we can write 2q(2p − 3)

Check. Expand the answer to see if you get the original back: 2q × 2p = 4pq; 2q × (−3) = −6q. Together: 4pq − 6q. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 12xy, 8y
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 12: 1, 2, 3, 4, 6, 12
   Factors of 8: 1, 2, 4, 8
   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).
   x: it is missing from at least one term (a missing letter has power 0), so it is not common.
   y: the powers in the terms are 1, 1. It is in every term and the lowest power is 1, so y is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 4y
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.
   12xy ÷ 4y = 3x
   8y ÷ 4y = 2
   So we can write 4y(3x + 2)

Check. Expand the answer to see if you get the original back: 4y × 3x = 12xy; 4y × 2 = 8y. Together: 12xy + 8y. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 5mn, −10n
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 10: 1, 2, 5, 10
   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).
   m: it is missing from at least one term (a missing letter has power 0), so it is not common.
   n: the powers in the terms are 1, 1. It is in every term and the lowest power is 1, so n is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 5n
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.
   5mn ÷ 5n = m
   −10n ÷ 5n = −2
   So we can write 5n(m − 2)

Check. Expand the answer to see if you get the original back: 5n × m = 5mn; 5n × (−2) = −10n. Together: 5mn − 10n. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 4x, 8y, 12
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 8: 1, 2, 4, 8
   Factors of 12: 1, 2, 3, 4, 6, 12
   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).
   x: it is missing from at least one term (a missing letter has power 0), so it is not common.
   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 = 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.
   4x ÷ 4 = x
   8y ÷ 4 = 2y
   12 ÷ 4 = 3
   So we can write 4(x + 2y + 3)

Check. Expand the answer to see if you get the original back: 4 × x = 4x; 4 × 2y = 8y; 4 × 3 = 12. Together: 4x + 8y + 12. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 10a, −15b, 5
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 10: 1, 2, 5, 10
   Factors of 15: 1, 3, 5, 15
   Factors of 5: 1, 5
   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).
   a: it is missing from at least one term (a missing letter has power 0), so it is not common.
   b: 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.
   10a ÷ 5 = 2a
   −15b ÷ 5 = −3b
   5 ÷ 5 = 1
   So we can write 5(2a − 3b + 1)

Check. Expand the answer to see if you get the original back: 5 × 2a = 10a; 5 × (−3b) = −15b; 5 × 1 = 5. Together: 10a − 15b + 5. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 3m², 6m, 9
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
   Factors of 9: 1, 3, 9
   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).
   m: 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.
   3m² ÷ 3 = m²
   6m ÷ 3 = 2m
   9 ÷ 3 = 3
   So we can write 3(m² + 2m + 3)

Check. Expand the answer to see if you get the original back: 3 × m² = 3m²; 3 × 2m = 6m; 3 × 3 = 9. Together: 3m² + 6m + 9. ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: x³, 5x²
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 5: 1, 5
   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 3, 2. It is in every term and the lowest power is 2, so x² is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = x²
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.
   x³ ÷ x² = x
   5x² ÷ x² = 5
   So we can write x²(x + 5)

Check. Expand the answer to see if you get the original back: x² × x = x³; x² × 5 = 5x². Together: x³ + 5x². ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 4x³y, −6x²y²
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 6: 1, 2, 3, 6
   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: the powers in the terms are 3, 2. It is in every term and the lowest power is 2, so x² is common.
   y: the powers in the terms are 1, 2. It is in every term and the lowest power is 1, so y is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 2x²y
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.
   4x³y ÷ 2x²y = 2x
   −6x²y² ÷ 2x²y = −3y
   So we can write 2x²y(2x − 3y)

Check. Expand the answer to see if you get the original back: 2x²y × 2x = 4x³y; 2x²y × (−3y) = −6x²y². Together: 4x³y − 6x²y². ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 12a³, 18a²b, −6ab²
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 12: 1, 2, 3, 4, 6, 12
   Factors of 18: 1, 2, 3, 6, 9, 18
   Factors of 6: 1, 2, 3, 6
   The biggest number that is in every list is 6.
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: the powers in the terms are 3, 2, 1. It is in every term and the lowest power is 1, so a is common.
   b: 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 = 6a
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.
   12a³ ÷ 6a = 2a²
   18a²b ÷ 6a = 3ab
   −6ab² ÷ 6a = −b²
   So we can write 6a(2a² + 3ab − b²)

Check. Expand the answer to see if you get the original back: 6a × 2a² = 12a³; 6a × 3ab = 18a²b; 6a × (−b²) = −6ab². Together: 12a³ + 18a²b − 6ab². ✓

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 with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

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: 15p²q, −10pq², 5pq
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 15: 1, 3, 5, 15
   Factors of 10: 1, 2, 5, 10
   Factors of 5: 1, 5
   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).
   p: the powers in the terms are 2, 1, 1. It is in every term and the lowest power is 1, so p is common.
   q: the powers in the terms are 1, 2, 1. It is in every term and the lowest power is 1, so q is common.
4. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 5pq
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.
   15p²q ÷ 5pq = 3p
   −10pq² ÷ 5pq = −2q
   5pq ÷ 5pq = 1
   So we can write 5pq(3p − 2q + 1)

Check. Expand the answer to see if you get the original back: 5pq × 3p = 15p²q; 5pq × (−2q) = −10pq²; 5pq × 1 = 5pq. Together: 15p²q − 10pq² + 5pq. ✓

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).

Method. Look for a common factor that is a whole bracket.

1. Spot the repeated bracket. The same bracket appears in both parts of the expression.
   the common bracket is (x + 4)
2. Take it out as a common factor. Write the common bracket once. In the other bracket put what is left from each part.
   What is left: 3 and y
3. Write the answer.
   (x + 4)(3 + y)

Check. Expand both the question and the answer: they give the same expression. Expanding the answer: xy + 3x + 4y + 12. ✓

Watch out. Do not expand first.

Method. Look for a common factor that is a whole bracket.

1. Spot the repeated bracket. The same bracket appears in both parts of the expression.
   the common bracket is (x − 2)
2. Take it out as a common factor. Write the common bracket once. In the other bracket put what is left from each part.
   What is left: x and 5
3. Write the answer.
   (x − 2)(x + 5)

Check. Expand both the question and the answer: they give the same expression. Expanding the answer: x² + 3x − 10. ✓

Watch out. Do not expand first.

Method. Look for a common factor that is a whole bracket.

1. Spot the repeated bracket. The same bracket appears in both parts of the expression.
   the common bracket is (b + 3)
2. Take it out as a common factor. Write the common bracket once. In the other bracket put what is left from each part.
   What is left: a and (−2)
3. Write the answer.
   (b + 3)(a − 2)

Check. Expand both the question and the answer: they give the same expression. Expanding the answer: ab + 3a − 2b − 6. ✓

Watch out. Do not expand first.

Method. Width = area ÷ length. Factorising the area shows the width.

1. Say what we know. Area = 6x² + 9x and length = 3x.
   area = length × width
2. 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: 6x², 9x
3. 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 6: 1, 2, 3, 6
   Factors of 9: 1, 3, 9
   The biggest number that is in every list is 3.
4. 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.
5. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 3x
6. 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.
   6x² ÷ 3x = 2x
   9x ÷ 3x = 3
   So we can write 3x(2x + 3)
7. Read off the width. The area is 3x × (2x + 3) and the length is 3x, so the width is the bracket.
   width = 2x + 3
8. Find the perimeter. Perimeter = 2 × (length + width). Collect like terms.
   2(3x + 2x + 3) = 2(5x + 3) = 10x + 6

Check. Area check: 3x(2x + 3) = 6x² + 9x. ✓ Perimeter check with x = 1: length 3, width 5, so 2(3 + 5) = 16 and 10 + 6 = 16. ✓

Method. Width = area ÷ length. Factorising the area shows the width.

1. Say what we know. Area = 6x² + 9x and length = 3x.
   area = length × width
2. 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: 6x², 9x
3. 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 6: 1, 2, 3, 6
   Factors of 9: 1, 3, 9
   The biggest number that is in every list is 3.
4. 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.
5. Write the highest common factor. Multiply the number part and the common letters together.
   HCF = 3x
6. 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.
   6x² ÷ 3x = 2x
   9x ÷ 3x = 3
   So we can write 3x(2x + 3)
7. Read off the width. The area is 3x × (2x + 3) and the length is 3x, so the width is the bracket.
   width = 2x + 3
8. Find the perimeter. Perimeter = 2 × (length + width). Collect like terms.
   2(3x + 2x + 3) = 2(5x + 3) = 10x + 6

Check. Area check: 3x(2x + 3) = 6x² + 9x. ✓ Perimeter check with x = 1: length 3, width 5, so 2(3 + 5) = 16 and 10 + 6 = 16. ✓