GCSE Maths Algebra Basics (Foundation): check-up and exam-style questions
Part of the GCSE Maths algebra basics chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Exercise 7Z Check-up (non-calculator)
Non-calculatorKey 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).
- 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.)+5a, +3b, −2a, +4b, −6, +9
- 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.a terms: +5a, −2a
add the coefficients: 5 − 2 = 3 so they give 3a
b terms: +3b, +4b
add the coefficients: 3 + 4 = 7 so they give 7b
number terms: −6, +9
add the coefficients: −6 + 9 = 3 so they give 3 - Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.3a + 7b + 3
Check. Test with numbers. Put a = 3, b = 4 into the original and into your answer. The original gives 40 and the answer gives 40. 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. Number first, then letters in order, repeated letters as a power.
- Collect the number and each letter. The number is 5, there is one a and two bs.5, a, b × b = b2
- Write them together. 5ab2
Check. With a = 2, b = 3: 2 × 3 × 3 × 5 = 90 and 5 × 2 × 9 = 90. ✓
Method. A division is written as a fraction.
- Put y on top and 4 on the bottom. y4
- Check the meaning. The fraction bar means divide.y over 4 = y ÷ 4
Check. With y = 20: 20 ÷ 4 = 5. ✓
Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.3x2 × 5x4
- Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x2 × x3 = (x × x) × (x × x × x) = x5 (the law am × an = am+n). A letter written on its own has power 1.Numbers: 3 × 5 = 15
x: add the powers 2 + 4 = 6
So the top is 15x6 - Write the answer. Put the number first, then the letters.15x6
Check. Test with numbers: put x = 2 into the question and into the answer. The question gives 960 and the answer gives 960. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.12a5 ÷ 3a2
- Now divide: divide the numbers, and for each letter subtract the powers. Dividing by a term is the same as cancelling. x5 ÷ x2 = (x × x × x × x × x) ÷ (x × x) = x3 (the law am ÷ an = am−n). Divide the numbers as you would divide any numbers.Numbers: 12 ÷ 3 = 4
a: subtract the powers 5 − 2 = 3 - Write the answer. Put the number first, then the letters.4a3
Check. Test with numbers: put a = 2 into the question and into the answer. The question gives 32 and the answer gives 32. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.(2y3)2
- Deal with any power outside a bracket first. A power outside a bracket applies to everything inside it: the number and every letter. For example (2x3)2 means 2x3 × 2x3. The number is raised to the power, and each letter's own power is multiplied by it (the law (am)n = amn).(2y3)2 → number: 22 = 4; y: 3 × 2 = 6 → 4y6
- Write the answer. Put the number first, then the letters.4y6
Check. Test with numbers: put y = 2 into the question and into the answer. The question gives 256 and the answer gives 256. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.
Method. To expand a bracket, multiply every term inside the bracket by the term outside it.
- Say what the bracket means. A number or letter written next to a bracket means multiply. So 4(2x − 3) means 4 multiplied by everything inside the bracket, not just the first term.4(2x − 3): 4 is multiplied by each of the terms +2x, −3
- Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). Watch the signs: Use the sign rules: a positive times a positive is positive, a negative times a negative is positive, and a positive times a negative (either way round) is negative.4 × 2x = 8x
4 × (−3) = −12 - Put the answers together. The results join up as a single expression. Each keeps its own sign.8x − 12
Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 4 and the answer gives 4. 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. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.
Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.
Method. To expand a bracket, multiply every term inside the bracket by the term outside it.
- Say what the bracket means. A number or letter written next to a bracket means multiply. So x(x + 6) means x multiplied by everything inside the bracket, not just the first term.x(x + 6) = x × x + x × 6
- Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). All the signs are positive here, so every answer is positive.x × x = x2
x × 6 = 6x - Put the answers together. The results join up as a single expression. Each keeps its own sign.x2 + 6x
Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 16 and the answer gives 16. 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. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.
Method. Expand every bracket first (keeping each sign with its term), then collect the like terms.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.3(x + 2) − 2(x − 1)
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front. A minus in front of a bracket means you multiply by a negative number, so every sign inside the bracket flips.Bracket: 3(x + 2)
3 × x = 3x
3 × 2 = 6
Bracket: −2(x − 1)
(−2) × x = −2x
(−2) × (−1) = 2 - Write all the pieces in one line. Now there are no brackets, only terms.3x + 6 − 2x + 2
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x terms: +3x, −2x
add the coefficients: 3 − 2 = 1 so they give x
number terms: +6, +2
add the coefficients: 6 + 2 = 8 so they give 8
Answer: x + 8
Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 10 and the answer gives 10. 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. A minus sign in front of a bracket changes the sign of every term inside it. −2(x − 5) = −2x + 10. Many mistakes are made with the last term.
Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.
Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.
- Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
First: x × x = x2× x −5 x x2 −5x 3 3x −15
Outer: x × (−5) = −5x
Inner: 3 × x = 3x
Last: 3 × (−5) = −15 - Write down all the pieces. Add together the results of all the multiplications, each with its own sign.x2 − 5x + 3x − 15
- Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.x2 terms: +x2 (only one of this type, so it stays as it is)
x terms: −5x, +3x
add the coefficients: −5 + 3 = −2 so they give −2x
number terms: −15 (only one of this type, so it stays as it is)
Answer: x2 − 2x − 15
Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives −12 and the answer gives −12. 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. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.
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.
- 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: 8x2, 12x
- 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. - 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.
- Write the highest common factor. Multiply the number part and the common letters together.HCF = 4x
- 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.8x2 ÷ 4x = 2x
12x ÷ 4x = 3
So we can write 4x(2x + 3)
Check. Expand the answer to see if you get the original back: 4x × 2x = 8x2; 4x × 3 = 12x. Together: 8x2 + 12x. ✓
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.
- 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.
- 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 ✗
−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. ✓ - 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 − 6)(x − 4)
Check. Expand the answer: (x − 6)(x − 4) 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 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.
- 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 − 81: x2 is a square and 81 is a square, with a minus between them. ✓
- 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)
- 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
81 = (9)2, so b = 9 - Put a and b into the rule. One bracket has a minus, the other has a plus.(x − 9)(x + 9)
Check. Expand to check. (x − 9)(x + 9): x × x = x2; x × 9 = 9x and (−9) × x = −9x (these two cancel); (−9) × 9 = −81. Total: x2 − 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.
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.
- 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 − 16: 9y2 is a square and 16 is a square, with a minus between them. ✓
- 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)
- 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
16 = (4)2, so b = 4 - Put a and b into the rule. One bracket has a minus, the other has a plus.(3y − 4)(3y + 4)
Check. Expand to check. (3y − 4)(3y + 4): 3y × 3y = 9y2; 3y × 4 = 12y and (−4) × 3y = −12y (these two cancel); (−4) × 4 = −16. Total: 9y2 − 16. ✓
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. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 2(x + 4) = 2x + 8
Right side: 2x + 8 = 2x + 8 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)2x + 8 and 2x + 8 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 12, right = 12; x = 5: left = 18, right = 18. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Total 29 marks. Allow about 29 minutes.
Exercise 7X Exam-style questions
Non-calculatorKey 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).
- 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.)+6a, +3b, −2a, +5b
- 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.a terms: +6a, −2a
add the coefficients: 6 − 2 = 4 so they give 4a
b terms: +3b, +5b
add the coefficients: 3 + 5 = 8 so they give 8b - Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.4a + 8b
Check. Test with numbers. Put a = 3, b = 4 into the original and into your answer. The original gives 44 and the answer gives 44. 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. Expand every bracket first (keeping each sign with its term), then collect the like terms.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.4(x + 3) − 2(x − 1)
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front. A minus in front of a bracket means you multiply by a negative number, so every sign inside the bracket flips.Bracket: 4(x + 3)
4 × x = 4x
4 × 3 = 12
Bracket: −2(x − 1)
(−2) × x = −2x
(−2) × (−1) = 2 - Write all the pieces in one line. Now there are no brackets, only terms.4x + 12 − 2x + 2
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x terms: +4x, −2x
add the coefficients: 4 − 2 = 2 so they give 2x
number terms: +12, +2
add the coefficients: 12 + 2 = 14 so they give 14
Answer: 2x + 14
Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 18 and the answer gives 18. 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. A minus sign in front of a bracket changes the sign of every term inside it. −2(x − 5) = −2x + 10. Many mistakes are made with the last term.
Method. Cost of several items = number × price of one. Add the two costs.
- Pens. 4 pens at p pence each.4 × p = 4p
- Books. 3 books at b pence each.3 × b = 3b
- Add. 4p + 3b pence
Check. If a pen is 50p and a book is 300p: 200 + 900 = 1100p, and 4 × 50 + 3 × 300 = 1100. ✓
Watch out. Marks: 1 for 4p, 1 for 3b.
Method. Change = amount paid − total cost. Use the same units: £10 = 1000 pence.
- Change £10 into pence. £1 = 100p.£10 = 1000p
- Subtract the total cost. Put the cost in a bracket, because it is all taken away.1000 − (4p + 3b)
- Remove the bracket. The minus changes both signs.1000 − 4p − 3b
Check. If p = 50, b = 300: 1000 − 200 − 900 = −100, so the money is not enough. With p = 50, b = 200: 1000 − 200 − 600 = 200. ✓
Watch out. Marks: 1 for 1000 and 1 for the correct subtraction.
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.
- 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
- 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. - 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.
- Write the highest common factor. Multiply the number part and the common letters together.HCF = 3x
- 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 binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.
Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.
- Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
First: x × x = x2× x −3 x x2 −3x 5 5x −15
Outer: x × (−3) = −3x
Inner: 5 × x = 5x
Last: 5 × (−3) = −15 - Write down all the pieces. Add together the results of all the multiplications, each with its own sign.x2 − 3x + 5x − 15
- Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.x2 terms: +x2 (only one of this type, so it stays as it is)
x terms: −3x, +5x
add the coefficients: −3 + 5 = 2 so they give 2x
number terms: −15 (only one of this type, so it stays as it is)
Answer: x2 + 2x − 15
Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 0 and the answer gives 0. 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. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.
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.
- 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.
- 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. ✓ - 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. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.p3 × p4
- Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x2 × x3 = (x × x) × (x × x × x) = x5 (the law am × an = am+n). A letter written on its own has power 1.Numbers: 1 × 1 = 1
p: add the powers 3 + 4 = 7
So the top is p7 - Write the answer. Put the number first, then the letters.p7
Check. Test with numbers: put p = 2 into the question and into the answer. The question gives 128 and the answer gives 128. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.y8 ÷ y2
- Now divide: divide the numbers, and for each letter subtract the powers. Dividing by a term is the same as cancelling. x5 ÷ x2 = (x × x × x × x × x) ÷ (x × x) = x3 (the law am ÷ an = am−n). Divide the numbers as you would divide any numbers.Numbers: 1 ÷ 1 = 1
y: subtract the powers 8 − 2 = 6 - Write the answer. Put the number first, then the letters.y6
Check. Test with numbers: put y = 2 into the question and into the answer. The question gives 64 and the answer gives 64. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x3 = x × x × x. The coefficient is the number in front of the letters.
Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.
- Write the question. We are going to simplify one product or quotient of terms.(m2)5
- Deal with any power outside a bracket first. A power outside a bracket applies to everything inside it: the number and every letter. For example (2x3)2 means 2x3 × 2x3. The number is raised to the power, and each letter's own power is multiplied by it (the law (am)n = amn).(m2)5 → m: 2 × 5 = 10 → m10
- Write the answer. Put the number first, then the letters.m10
Check. Test with numbers: put m = 2 into the question and into the answer. The question gives 1024 and the answer gives 1024. They are the same. ✓
Watch out. Add the powers when you multiply (x2 × x3 = x5, not x6). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x2, the 3 and 2 are multiplied, not added.
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.
- 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.25x2 − 4: 25x2 is a square and 4 is a square, with a minus between them. ✓
- 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)
- 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.25x2 = (5x)2, so a = 5x
4 = (2)2, so b = 2 - Put a and b into the rule. One bracket has a minus, the other has a plus.(5x − 2)(5x + 2)
Check. Expand to check. (5x − 2)(5x + 2): 5x × 5x = 25x2; 5x × 2 = 10x and (−2) × 5x = −10x (these two cancel); (−2) × 2 = −4. Total: 25x2 − 4. ✓
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. Expand the first product, then add 16 and compare with the right side.
- Expand (x + 4)(x − 4). Four pieces: x × x, x × −4, 4 × x, 4 × −4.(x + 4)(x − 4) = x2 − 4x + 4x − 16 = x2 − 16
- Add 16. The question has (x + 4)(x − 4) + 16.(x + 4)(x − 4) + 16 = x2 − 16 + 16 = x2
- Conclude. The left side simplifies to x2, so (x + 4)(x − 4) + 16 ≡ x2. ✓
Check. Test x = 6: (10)(2) + 16 = 36 = 62. ✓
Watch out. Marks: 1 for expanding, 1 for simplifying to x2, 1 for the conclusion.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 4x + 1 = 4x + 1
Right side: 21 = 21 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.4x + 1 and 21 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 0: left side = 1, right side = 21. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 5.
Check. Put x = 5 into both sides: left = 21 and right = 21, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 3(x + 1) = 3x + 3
Right side: 3x + 3 = 3x + 3 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)3x + 3 and 3x + 3 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 9, right = 9; x = 5: left = 18, right = 18. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: (x + 2)(x − 2) = x2 − 4
Right side: x2 − 4 = x2 − 4 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)x2 − 4 and x2 − 4 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 0, right = 0; x = 5: left = 21, right = 21. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the four consecutive odd numbers. An odd number is 2n + 1, and consecutive odd numbers go up in 2s.2n + 1, 2n + 3, 2n + 5, 2n + 7
- Add them and simplify. (2n + 1) + (2n + 3) + (2n + 5) + (2n + 7) = 8n + 16
- Factorise. The common factor of 8 and 16 is 8.8n + 16 = 8(n + 2)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n + 2 is an integer, so 8(n + 2) is a multiple of 8. So the sum of any four consecutive odd numbers is a multiple of 8. ✓
Check. Test with 3, 5, 7, 9: sum 24 = 8 × 3. n = 1: 8(1 + 2) = 24. ✓
Watch out. Marks: 1 for the four numbers, 1 for adding and simplifying, 1 for factorising and the conclusion.
Method. Area = length × width, so factorise the area and match one bracket to the width.
- Write what we know. The area is a product of length and width.length × (x + 2) = x2 + 5x + 6
- 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. ✓ - Match the brackets. The factorised area is (x + 2)(x + 3). The width is (x + 2), so the length is the other bracket.length = (x + 3) cm
Check. (x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6. ✓
Watch out. Marks: 1 for a correct method, 1 for the correct factorisation.