Maths · 2 Algebra

2.2 Algebraic manipulation

Before you start

Adding, subtracting and multiplying negative numbers, e.g. $- 3 \times - 4 = 12$ and $- 2 - 5 = - 7$ .
Multiplying powers of the same letter: $x \times x = x^{2}$ and $x^{2} \times x = x^{3}$ .
Finding the highest common factor (HCF) of two or three numbers, e.g. the HCF of $12$ and $18$ is $6$ .
Substituting a number into an expression, e.g. $2 x^{2} - 1 = 17$ when $x = 3$ .

Syllabus objectives covered (4)

Simplify expressions by collecting like terms.
Expand products of algebraic expressions.
Factorise by extracting common factors.
Factorise expressions of the form: • ax + bx + kay + kby • a²x² − b²y² • a² + 2ab + b² • ax² + bx + c • ax³ + bx² + cx.

Part 1 of 8

Like terms: what can be collected

A term is a number, a letter, or numbers and letters multiplied together: $7$ , $- 3 x$ , $4 a b$ , $5 a^{2}$ . The sign in front of a term belongs to it, so in $6 - 2 x$ the second term is $- 2 x$ , not $2 x$ .
Two terms are like terms when they have exactly the same letters raised to exactly the same powers. $3 a b$ and $- 5 b a$ are like terms, because $b a$ is the same as $a b$ . $2 a^{2}$ and $2 a$ are not: one is $a \times a$ , the other is just $a$ .
Collecting like terms means adding their coefficients and leaving the letter part alone. $7 a^{2} + 3 a^{2}$ is seven lots of $a^{2}$ plus three lots of $a^{2}$ , which is ten lots of $a^{2}$ . The power does not change, so the answer is not $10 a^{4}$ .
$7 a^{2} + 3 a^{2} = 10 a^{2}$
To simplify a longer expression, sort the terms by type and carry each sign with its term. In $5 x^{2} - 2 x y + 6 - 3 x^{2} + 7 x y - 9$ the $x^{2}$ terms are $5 x^{2}$ and $- 3 x^{2}$ , the $x y$ terms are $- 2 x y$ and $+ 7 x y$ , and the numbers are $6$ and $- 9$ .
$(5 x^{2} - 3 x^{2}) + (- 2 x y + 7 x y) + (6 - 9) = 2 x^{2} + 5 x y - 3$
Your answer is fully simplified when no two of its terms are alike. A quick check: three types of term went in, so at most three terms can come out.
Two cases need care. $x^{2} + x$ stays as it is, because the terms are unlike. A pair such as $4 x - 4 x$ gives $0$ , so that type of term disappears from the answer completely; do not write $0 x$ .

Part 2 of 8

Expanding a single bracket

To expand a bracket, multiply every term inside it by the term outside. The bracket tells you that the outside term applies to everything within it.
$3 (2 x - 5) = 6 x - 15$
When the outside term contains a letter, multiply the letters with the index rules: $a \times a = a^{2}$ , and different letters are written side by side.
$4 a (3 a - 2 b) = 12 a^{2} - 8 a b$
A negative outside the bracket multiplies every term, signs included. In $- 2 (3 x - 5)$ , the $- 2$ times the $- 5$ gives $+ 10$ , because a negative times a negative is positive.
$- 2 (3 x - 5) = - 6 x + 10$
Expand and simplify means do both jobs: multiply out every bracket, then collect like terms.
$5 (x + 2) - 3 (x - 4) = 5 x + 10 - 3 x + 12 = 2 x + 22$
The trap in that line is the $+ 12$ . The $- 3$ multiplies the $- 4$ , so the sign changes. Write the expansion of each bracket in full before you collect anything, and every sign is in front of you.

Part 3 of 8

Two brackets: every term times every term

With two brackets, each term in the first bracket multiplies each term in the second. Two terms times two terms gives four products before you simplify.
A rectangle shows why. A rectangle $(x + 3)$ long and $(x + 2)$ wide splits into four smaller rectangles with areas $x^{2}$ , $3 x$ , $2 x$ and $6$ . The whole area is the sum of the four pieces.
$(x + 3) (x + 2) = x^{2} + 2 x + 3 x + 6 = x^{2} + 5 x + 6$
graph diagram — not drawn yet
A rectangle divided into a 2-by-2 grid of smaller rectangles. The top edge is split into a long piece labelled x and a short piece labelled 3; the left edge is split into a long piece labelled x and a short piece labelled 2. The four regions are labelled x² (top left), 3x (top right), 2x (bottom left) and 6 (bottom right).
Each region is one of the four products; added together they give $(x + 3) (x + 2) = x^{2} + 5 x + 6$ .
Signs travel with their terms, exactly as before. In $(2 x - 3) (x + 4)$ the four products are $2 x \times x$ , $2 x \times 4$ , $- 3 \times x$ and $- 3 \times 4$ .
$(2 x - 3) (x + 4) = 2 x^{2} + 8 x - 3 x - 12 = 2 x^{2} + 5 x - 12$
Two letters work the same way. Here the two middle products are both $a b$ terms, so they collect.
$(2 a + b) (a - 3 b) = 2 a^{2} - 6 a b + a b - 3 b^{2} = 2 a^{2} - 5 a b - 3 b^{2}$
A squared bracket means the bracket times itself: ${(2 x + 5)}^{2} = (2 x + 5) (2 x + 5)$ . Write it as two brackets before you start. Both middle products are $10 x$ , so the middle term is $20 x$ .
${(2 x + 5)}^{2} = 4 x^{2} + 10 x + 10 x + 25 = 4 x^{2} + 20 x + 25$
Notice the pattern in that answer: square the first term, square the last term, and the middle term is twice their product. You will use this pattern backwards when you factorise perfect squares.
Check any expansion by substituting a small number. With $x = 1$ , $(2 x - 3) (x + 4) = (- 1) (5) = - 5$ , and $2 x^{2} + 5 x - 12 = 2 + 5 - 12 = - 5$ . If the two values disagree, the expansion is wrong.

Part 4 of 8

Three brackets: two at a time

You can only multiply two things at once, so expand any two of the brackets first, then multiply the result by the third.
Take $(x + 2) (x - 1) (x + 3)$ . Start with the first two brackets.
$(x + 2) (x - 1) = x^{2} - x + 2 x - 2 = x^{2} + x - 2$
Now multiply each of the three terms of that quadratic by each of the two terms of $(x + 3)$ . Three times two gives six products.
$(x^{2} + x - 2) (x + 3) = x^{3} + 3 x^{2} + x^{2} + 3 x - 2 x - 6$
Collect like terms to finish.
$x^{3} + 4 x^{2} + x - 6$
Keep the quadratic from the first step inside a bracket while you multiply, so that no term of it gets left behind. If you count fewer than six products, you have missed one.
Check with a number that does not make any bracket zero. With $x = 2$ : $(4) (1) (5) = 20$ , and $8 + 16 + 2 - 6 = 20$ . They agree.

Part 5 of 8

Factorising by taking out a common factor

Factorising is expanding in reverse. Expanding takes $3 (2 x + 5)$ to $6 x + 15$ ; factorising takes $6 x + 15$ back to $3 (2 x + 5)$ .
Find the highest common factor of every term. For the numbers, use their HCF. For each letter, take the lowest power that appears in every term. A letter that is missing from any one term cannot be taken out.
Take $12 x^{2} y + 18 x y^{2}$ . The HCF of $12$ and $18$ is $6$ . The lowest power of $x$ in the two terms is $x$ , and the lowest power of $y$ is $y$ . So the common factor is $6 x y$ .
Divide each term by the common factor to find what goes inside the bracket: $12 x^{2} y \div 6 x y = 2 x$ and $18 x y^{2} \div 6 x y = 3 y$ .
$12 x^{2} y + 18 x y^{2} = 6 x y (2 x + 3 y)$
Factorise means factorise fully. $12 x^{2} + 18 x = 3 x (4 x + 6)$ is true, but $4 x + 6$ still has a factor of $2$ in it. The full answer is $6 x (2 x + 3)$ . Before you move on, look inside the bracket and ask whether anything still divides every term.
When a term is the whole common factor, a $1$ is left behind. In $5 a^{2} + a$ the common factor is $a$ , and $a \div a = 1$ , so the answer is $a (5 a + 1)$ . If you leave out the $1$ you get $a (5 a)$ , which expands to $5 a^{2}$ only.
Every factorisation can be checked by expanding it. If it does not expand back to the expression you started with, it is wrong.

Part 6 of 8

Four terms: factorising by grouping

An expression of the form $a x + b x + k a y + k b y$ has four terms, usually with no factor common to all of them. Split it into two pairs and factorise each pair on its own.
Take $3 p x + 2 q x + 12 p y + 8 q y$ . The first pair has the common factor $x$ . The second pair has the common factor $4 y$ .
$3 p x + 2 q x + 12 p y + 8 q y = x (3 p + 2 q) + 4 y (3 p + 2 q)$
Now the same bracket, $(3 p + 2 q)$ , appears in both parts. That bracket is a common factor of the whole expression, so take it out exactly as you would take out a single letter.
$(3 p + 2 q) (x + 4 y)$
When the third term is negative, take a negative factor out of the second pair so that the brackets match. In $p x + q x - 2 p y - 2 q y$ , take out $- 2 y$ : $- 2 p y - 2 q y = - 2 y (p + q)$ .
$p x + q x - 2 p y - 2 q y = x (p + q) - 2 y (p + q) = (p + q) (x - 2 y)$
The two brackets must be identical, signs included, before you can take them out. If they are not, check the sign of the factor you took from the second pair, or try pairing the first term with the third instead.
Check by expanding: $(p + q) (x - 2 y) = p x - 2 p y + q x - 2 q y$ , which is the original four terms in a different order.

Part 7 of 8

Two patterns: difference of two squares and perfect squares

Expand $(a x - b y) (a x + b y)$ and look at the middle. The products $+ a b x y$ and $- a b x y$ cancel, so only two terms are left.
$(a x - b y) (a x + b y) = a^{2} x^{2} + a b x y - a b x y - b^{2} y^{2} = a^{2} x^{2} - b^{2} y^{2}$
Read that backwards: a square minus a square factorises into a minus bracket times a plus bracket. Find the square root of each term to see what goes in the brackets.
$16 x^{2} - 49 y^{2} = (4 x - 7 y) (4 x + 7 y)$
You can recognise it by three features: exactly two terms, a minus sign between them, and both terms perfect squares (square numbers, with even powers of the letters).
Take out any common factor first, or the squares stay hidden. $12 x^{2} - 27$ does not look like two squares, but $12 x^{2} - 27 = 3 (4 x^{2} - 9)$ , and $4 x^{2} - 9$ is ${(2 x)}^{2} - 3^{2}$ .
$12 x^{2} - 27 = 3 (2 x - 3) (2 x + 3)$
A sum of two squares is different. $x^{2} + 9$ does not factorise: $(x + 3) (x - 3)$ gives $x^{2} - 9$ , and ${(x + 3)}^{2}$ gives $x^{2} + 6 x + 9$ .
The second pattern is the perfect square. It is the squared-bracket expansion from earlier, read backwards.
$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$
To test a three-term expression, find the square roots of the first and last terms, then check that the middle term is twice their product. For $x^{2} + 14 x + 49$ the roots are $x$ and $7$ , and $2 \times x \times 7 = 14 x$ , so $x^{2} + 14 x + 49 = {(x + 7)}^{2}$ .
A negative middle term gives a minus in the bracket. For $4 x^{2} - 12 x + 9$ the roots are $2 x$ and $3$ , and $2 \times 2 x \times 3 = 12 x$ , so $4 x^{2} - 12 x + 9 = {(2 x - 3)}^{2}$ . If the middle term fails this test, the expression is not a perfect square; use the method in the next section instead.

Part 8 of 8

Three-term quadratics, and cubics with a common x

For $x^{2} + b x + c$ (where $a = 1$ ), you need two numbers that multiply to give $c$ and add to give $b$ . For $x^{2} + 2 x - 15$ : multiply to $- 15$ , add to $2$ . The pair is $5$ and $- 3$ .
$x^{2} + 2 x - 15 = (x + 5) (x - 3)$
The sign of $c$ tells you where to look. If $c$ is positive, both numbers have the same sign as $b$ . If $c$ is negative, the numbers have opposite signs, and the one further from zero has the sign of $b$ .
When $a$ is not $1$ , guessing brackets is slow. A reliable method is to split the middle term. For $2 x^{2} + 7 x + 3$ , work out $a \times c = 2 \times 3 = 6$ , then find two numbers that multiply to $6$ and add to $b = 7$ . They are $6$ and $1$ .
Write $7 x$ as $6 x + x$ . Now there are four terms, and you factorise them by grouping, exactly as in the section on four terms.
$2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3)$
Keep the $1$ in $1 (x + 3)$ . It shows the matching bracket and stops you losing the $+ 1$ from the final answer.
Check the middle term, because that is where errors hide: in $(2 x + 1) (x + 3)$ the middle products are $2 x \times 3 = 6 x$ and $1 \times x = x$ , which add to $7 x$ . Correct.
The cubic form $a x^{3} + b x^{2} + c x$ has no constant term, so every term contains $x$ . Take out $x$ (and any number factor) first. What is left in the bracket is a quadratic, which you factorise as above.
$x^{3} - 3 x^{2} - 10 x = x (x^{2} - 3 x - 10) = x (x - 5) (x + 2)$
This gives an order of checks for any factorise fully question. First, take out every common factor. Then count the terms left in the bracket: two terms, try the difference of two squares; three terms, try a perfect square, then the multiply-and-add method; four terms, try grouping. Last, look at each bracket in your answer and ask whether it factorises again.

Worked examples

Example 1. Expand and simplify $2 x (3 x - 4) - (x - 5) (x + 2)$ .

Show the working

Expand the single bracket first.
$2 x (3 x - 4) = 6 x^{2} - 8 x$
Expand the pair of brackets on its own, and keep the result inside a bracket. The minus sign in front applies to the whole product, not only to its first term.
$(x - 5) (x + 2) = x^{2} + 2 x - 5 x - 10 = x^{2} - 3 x - 10$
Now subtract the whole bracket. Subtracting a bracket changes the sign of every term in it.
$6 x^{2} - 8 x - (x^{2} - 3 x - 10) = 6 x^{2} - 8 x - x^{2} + 3 x + 10$
Collect like terms: $6 x^{2} - x^{2} = 5 x^{2}$ , $- 8 x + 3 x = - 5 x$ , and the number term is $+ 10$ .
$5 x^{2} - 5 x + 10$
Check with $x = 1$ . The original gives $2 (1) (- 1) - (- 4) (3) = - 2 + 12 = 10$ , and $5 - 5 + 10 = 10$ . They agree.

Answer: $5 x^{2} - 5 x + 10$

Example 2. Expand and simplify $(x - 3) (2 x + 1) (x + 4)$ .

Show the working

Choose two brackets to start with. $(x - 3) (x + 4)$ is the easiest pair, because both begin with $x$ .
$(x - 3) (x + 4) = x^{2} + 4 x - 3 x - 12 = x^{2} + x - 12$
Multiply that quadratic by the third bracket. Each of its three terms multiplies $2 x$ and then $1$ .
$(x^{2} + x - 12) (2 x + 1) = 2 x^{3} + x^{2} + 2 x^{2} + x - 24 x - 12$
Count the products: six, as expected for three terms times two. Now collect like terms.
$2 x^{3} + 3 x^{2} - 23 x - 12$
Check with $x = 2$ : $(2 - 3) (4 + 1) (2 + 4) = (- 1) (5) (6) = - 30$ , and $16 + 12 - 46 - 12 = - 30$ . They agree.

Answer: $2 x^{3} + 3 x^{2} - 23 x - 12$

Example 3. Factorise fully $6 x^{3} + 3 x^{2} - 30 x$ .

Show the working

Look for a common factor first. Every term contains $x$ , and $6$ , $3$ and $30$ share the factor $3$ . So take out $3 x$ .
$6 x^{3} + 3 x^{2} - 30 x = 3 x (2 x^{2} + x - 10)$
The bracket is a quadratic with $a = 2$ . Work out $a \times c = 2 \times (- 10) = - 20$ . You need two numbers that multiply to $- 20$ and add to $b = 1$ : they are $5$ and $- 4$ .
Split the middle term $x$ as $5 x - 4 x$ and group the four terms in pairs.
$2 x^{2} + 5 x - 4 x - 10 = x (2 x + 5) - 2 (2 x + 5) = (2 x + 5) (x - 2)$
Put the common factor back in front. Leaving it out would give a different expression.
$6 x^{3} + 3 x^{2} - 30 x = 3 x (2 x + 5) (x - 2)$
Check the quadratic's middle term: $2 x \times (- 2) + 5 \times x = - 4 x + 5 x = x$ . Correct. Neither bracket factorises again, so the answer is fully factorised.

Answer: $3 x (2 x + 5) (x - 2)$

Example 4. Factorise fully (a) $18 x^{2} - 50 y^{2}$ (b) $2 a x + 3 b x - 6 a y - 9 b y$ .

Show the working

(a) Two terms and a minus sign suggest a difference of two squares, but $18$ and $50$ are not square numbers. Take out the common factor $2$ first.
$18 x^{2} - 50 y^{2} = 2 (9 x^{2} - 25 y^{2})$
Now $9 x^{2} = {(3 x)}^{2}$ and $25 y^{2} = {(5 y)}^{2}$ , so the bracket is a difference of two squares.
$2 (9 x^{2} - 25 y^{2}) = 2 (3 x - 5 y) (3 x + 5 y)$
(b) Four terms with no factor common to all four: group them in pairs. The first pair shares $x$ .
$2 a x + 3 b x = x (2 a + 3 b)$
The second pair shares $3 y$ , but both terms are negative, so take out $- 3 y$ . That makes the bracket match the first one.
$- 6 a y - 9 b y = - 3 y (2 a + 3 b)$
Both parts now contain $(2 a + 3 b)$ , so take it out as a common factor.
$x (2 a + 3 b) - 3 y (2 a + 3 b) = (2 a + 3 b) (x - 3 y)$

Answer: (a) $2 (3 x - 5 y) (3 x + 5 y)$ (b) $(2 a + 3 b) (x - 3 y)$

Where people go wrong

Mistake: Writing ${(x + 5)}^{2} = x^{2} + 25$ .

Instead: Write it as $(x + 5) (x + 5)$ and expand all four products: $x^{2} + 5 x + 5 x + 25 = x^{2} + 10 x + 25$ .

Squaring a bracket multiplies the whole bracket by itself. The two middle products do not cancel. Test with $x = 1$ : ${(1 + 5)}^{2} = 36$ , but $1 + 25 = 26$ .

Mistake: Expanding $- 3 (x - 4)$ as $- 3 x - 12$ .

Instead: $- 3 (x - 4) = - 3 x + 12$ .

The $- 3$ multiplies the $- 4$ as well as the $x$ , and a negative times a negative is positive.

Mistake: Stopping at $12 x^{2} + 18 x = 3 x (4 x + 6)$ .

Instead: Take out the HCF of every part: $12 x^{2} + 18 x = 6 x (2 x + 3)$ .

$4 x + 6$ still has a common factor of $2$ , so the expression is not fully factorised, and Factorise means factorise fully.

Mistake: Combining unlike terms, e.g. $3 x^{2} + 2 x = 5 x^{3}$ .

Instead: $3 x^{2} + 2 x$ cannot be simplified. It can only be factorised, as $x (3 x + 2)$ .

$x^{2}$ and $x$ are different quantities, so their coefficients cannot be added. Test with $x = 2$ : $3 x^{2} + 2 x = 16$ , but $5 x^{3} = 40$ .

Mistake: Factorising $x^{2} + 9$ as $(x + 3) (x - 3)$ .

Instead: $x^{2} - 9 = (x - 3) (x + 3)$ , but $x^{2} + 9$ does not factorise.

Expanding $(x + 3) (x - 3)$ gives $x^{2} - 9$ . The difference of two squares needs a minus sign between the squares.

Check yourself

Simplify $5 m - 2 n - 3 m + 6 n - m$ .
Answer
$m$ terms: $5 m - 3 m - m = m$ . $n$ terms: $- 2 n + 6 n = 4 n$ . Answer: $m + 4 n$ .
Simplify $3 x^{2} + 4 x - 5 - x^{2} - 7 x + 2$ .
Answer
$3 x^{2} - x^{2} = 2 x^{2}$ , $4 x - 7 x = - 3 x$ , $- 5 + 2 = - 3$ . Answer: $2 x^{2} - 3 x - 3$ .
Expand and simplify $3 (2 x - 1) - 2 (x - 4)$ .
Answer
$6 x - 3 - 2 x + 8 = 4 x + 5$ . The $- 2$ times $- 4$ gives $+ 8$ .
Expand and simplify ${(2 x - 5)}^{2}$ .
Answer
$(2 x - 5) (2 x - 5) = 4 x^{2} - 10 x - 10 x + 25 = 4 x^{2} - 20 x + 25$ .
Factorise fully $12 a^{2} b - 18 a b^{2}$ .
Answer
The HCF is $6 a b$ , so $12 a^{2} b - 18 a b^{2} = 6 a b (2 a - 3 b)$ .
Factorise $a p + b p + 3 a q + 3 b q$ .
Answer
$p (a + b) + 3 q (a + b) = (a + b) (p + 3 q)$ .
Factorise $9 x^{2} + 12 x y + 4 y^{2}$ .
Answer
The square roots of the end terms are $3 x$ and $2 y$ , and $2 \times 3 x \times 2 y = 12 x y$ , so it is a perfect square: ${(3 x + 2 y)}^{2}$ .
Factorise fully $2 x^{3} - 5 x^{2} - 3 x$ .
Answer
Take out $x$ : $x (2 x^{2} - 5 x - 3)$ . Then $a \times c = - 6$ ; the numbers $- 6$ and $1$ multiply to $- 6$ and add to $- 5$ . So $x (2 x + 1) (x - 3)$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Simplify $6 a^{2} - 4 a b + 3 - 2 a^{2} - a b - 8$ . 2 marks No calculator

Mark scheme

B2 $4 a^{2} - 5 a b - 5$ ; B1 for two of the three terms correct in the final answer

(b) Expand and simplify $(3 x - 2) (x + 5)$ . 2 marks No calculator

Mark scheme

M1 at least three of $3 x^{2}$ , $15 x$ , $- 2 x$ , $- 10$ correct
A1 $3 x^{2} + 13 x - 10$ cao

(c) Expand and simplify $(x + 1) (x - 3) (2 x + 5)$ . 3 marks No calculator

Mark scheme

M1 correct expansion of any two brackets, e.g. $x^{2} - 2 x - 3$ or $2 x^{2} + 7 x + 5$ or $2 x^{2} - x - 15$ ; or, expanding all three brackets at once, at least six of the eight products correct
M1 their quadratic multiplied by the remaining bracket, with at least four of the six products correct; FT their quadratic; or all eight products of a direct expansion correct
A1 $2 x^{3} + x^{2} - 16 x - 15$ cao

Question 2. Factorise each expression fully.

(a) $50 x^{2} - 18 y^{2}$ 2 marks No calculator

Mark scheme

B2 $2 (5 x - 3 y) (5 x + 3 y)$ oe; B1 for $2 (25 x^{2} - 9 y^{2})$ or $(10 x - 6 y) (5 x + 3 y)$ or $(5 x - 3 y) (10 x + 6 y)$

(b) $3 x^{2} + 10 x - 8$ 2 marks No calculator

Mark scheme

M1 $(3 x + a) (x + b)$ where $a b = - 8$ or $a + 3 b = 10$ ; or the middle term split correctly, e.g. $3 x^{2} + 12 x - 2 x - 8$ oe
A1 $(3 x - 2) (x + 4)$ oe

(c) $4 x^{3} - 4 x^{2} - 24 x$ 3 marks No calculator

Mark scheme

B1 $4 x (x^{2} - x - 6)$ soi
M1 $(x - 3) (x + 2)$ seen, or a correct factorisation of their three-term quadratic
A1 $4 x (x - 3) (x + 2)$ cao

(d) $m x + n x + 4 m y + 4 n y$ 2 marks No calculator

Mark scheme

M1 $x (m + n) + 4 y (m + n)$ or $m (x + 4 y) + n (x + 4 y)$
A1 $(m + n) (x + 4 y)$ oe

Question 3. A square tile has side $(3 x + 2)$ cm, where $x > 0$ . A square of side $(x + 2)$ cm is cut from one corner of the tile. The area of the tile that is left is $A$ cm².

(a) By expanding brackets, show that $A = 8 x^{2} + 8 x$ . 2 marks No calculator

Mark scheme

M1 ${(3 x + 2)}^{2} - {(x + 2)}^{2}$ with both squares expanded correctly: $9 x^{2} + 12 x + 4$ and $x^{2} + 4 x + 4$ ; or the area split into rectangles, e.g. $2 x (3 x + 2) + 2 x (x + 2)$ , with each product expanded correctly
A1 every term of the second expansion subtracted, reaching $8 x^{2} + 8 x$ with no errors seen (or the rectangle areas added correctly); nfww

(b) By factorising ${(3 x + 2)}^{2} - {(x + 2)}^{2}$ as a difference of two squares, show that $A = 8 x (x + 1)$ . 2 marks No calculator

Mark scheme

M1 $[(3 x + 2) - (x + 2)] [(3 x + 2) + (x + 2)]$ seen oe
A1 $2 x (4 x + 4)$ simplified to $8 x (x + 1)$ with no errors seen

How this is examined

Factorising fully when it takes more than one step: the common factor first, then the bracket: a Factorise question on an expression such as $a x^{3} + b x^{2} + c x$ or $k (a^{2} x^{2} - b^{2} y^{2})$ , where the full answer has three factors The common factor taken out correctly, e.g. $5 x (x^{2} + x - 6)$ , then the bracket factorised, e.g. $5 x (x + 3) (x - 2)$ . An answer that stops after the first step has not factorised fully, so it cannot earn the final mark.
Expanding three brackets: two first, then the third: an Expand and simplify part with three linear brackets, whose answer is a cubic with four terms A correct quadratic from the first two brackets, e.g. $x^{2} + x - 6$ ; then all six products from multiplying by the third bracket; then the like terms collected into the final four-term cubic.
Factorising $a x^{2} + b x + c$ when $a$ is not 1: a two-mark Factorise on a three-term quadratic; the factors may then be needed in a later part of the same question Brackets of the right form, such as $(2 x + p) (x + q)$ , whose numbers give either the constant term or the middle term earn the method mark (so does a correct split of the middle term). Only the pair that gives both, e.g. $(2 x - 3) (x + 5)$ , earns the accuracy mark.
Proving a printed result by expanding: a Show (that) part in which the answer is given, e.g. an area that must equal a stated expression Every expansion written out in full, a bracket kept round anything being subtracted, and the working ending at exactly the printed expression. Writing the given answer with no expansions shown earns nothing, because nothing has been shown.
Recognising a difference of two squares or a perfect square on sight: a Factorise with two square terms and a minus sign, or three terms whose first and last terms are squares $(a x - b y) (a x + b y)$ with both square roots correct; for a perfect square, ${(a + b)}^{2}$ with the middle term checked as $2 a b$ .
Collecting like terms with the sign in front of each term: a one- or two-mark Simplify with several types of term Each type of term collected with its own sign, e.g. $- 2 x y - 3 x y = - 5 x y$ , and no like terms left in the final answer. One sign slip makes that term wrong, so full marks need every term right.

Words to know

Term: A single number, letter, or product of numbers and letters, together with the sign in front of it, e.g. $- 4 x y$ .
Coefficient: The number that multiplies the letters in a term. In $- 4 x y$ the coefficient is $- 4$ .
Like terms: Terms with exactly the same letters raised to exactly the same powers, such as $3 a^{2} b$ and $- a^{2} b$ .
Expand: Multiply out the brackets, so that the expression is written as a sum of terms.
Factorise: Write an expression as a product of factors. In an exam it means factorise fully, so no bracket can be factorised again.
Common factor: A number, letter or bracket that divides every term of an expression, e.g. $3 x$ in $6 x^{2} + 3 x$ .
Difference of two squares: An expression of the form $a^{2} - b^{2}$ , which factorises as $(a - b) (a + b)$ .
Perfect square: An expression of the form $a^{2} + 2 a b + b^{2}$ , which factorises as ${(a + b)}^{2}$ .

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