Simplifying, expanding & factorising (incl. quadratics) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

(a) Simplify $4a + 3b - a + 5b$ [1]

(b) Simplify $3x^2 \times 5x^4$ [1]

(a) $3a + 8b$ (accept $8b + 3a$)B1
(b) $15x^6$B1
(c) $10y - 15$B1
Answer: (a) $3a + 8b$ (b) $15x^6$ (c) $10y - 15$
Examiner tip: In (b) multiply the numbers but ADD the powers – $15x^8$ (multiplying the powers) gets no marks.

2Grade 4–5Non-calculator4 marks

(a) Factorise fully $12p^2 - 18pq$ [2]

(b) Expand and simplify $3(2x + 1) - 2(x - 4)$ [2]

(a) for a correct partial factorisation with two factors, e.g. $3p(4p - 6q)$, $6(2p^2 - 3pq)$ or $2p(6p - 9q)$B1
(a) (total B2) for the fully factorised answer $6p(2p - 3q)$B1
(b) for expanding both brackets with at least 3 of the 4 terms correct: $6x + 3 - 2x + 8$M1
$4x + 11$A1
Answer: (a) $6p(2p - 3q)$ (b) $4x + 11$
Examiner tip: 'Fully' means take out the HCF of the numbers AND the letters; and $-2 \times -4 = +8$, not $-8$.

3Grade 4–5Non-calculator4 marks

(a) Expand and simplify $(x + 5)(x - 3)$ [2]

(b) Factorise $x^2 - 9x + 20$ [2]

(a) for 4 terms with at least 3 correct, e.g. $x^2 - 3x + 5x - 15$M1
$x^2 + 2x - 15$A1
(b) for $(x \pm 4)(x \pm 5)$M1
$(x - 4)(x - 5)$ (either order)A1
Answer: (a) $x^2 + 2x - 15$ (b) $(x - 4)(x - 5)$
Examiner tip: For (b) you need two numbers that multiply to $+20$ and add to $-9$ – both must be negative.

4Grade 6–7HigherNon-calculator3 marks

(a) Factorise $6x^2 + 7x - 20$ [2]

(b) Factorise $49y^2 - 16$ [1]

(a) for $(3x \pm 4)(2x \pm 5)$, or for splitting the middle term correctly, e.g. $6x^2 + 15x - 8x - 20$, or for two brackets that expand to give two of the three terms correctlyM1
$(3x - 4)(2x + 5)$A1
(b) $(7y - 4)(7y + 4)$ (either order)B1
Answer: (a) $(3x - 4)(2x + 5)$ (b) $(7y - 4)(7y + 4)$
Examiner tip: Always expand your brackets back out to check the middle term – a sign swap gives $-7x$ and loses the A mark.

5Grade 6–7HigherNon-calculator3 marks

Expand and simplify $(2x - 1)(x + 3)(x - 4)$ [3]

for expanding any two of the brackets with at least 3 of 4 terms correct, e.g. $(x + 3)(x - 4) = x^2 - x - 12$M1
(dep) for a correct method to multiply their quadratic by the third bracket, giving 6 terms (or 8 terms if all three brackets are multiplied directly) with at least 4 correct, e.g. $2x^3 - 2x^2 - 24x - x^2 + x + 12$M1
$2x^3 - 3x^2 - 23x + 12$A1
Answer: $2x^3 - 3x^2 - 23x + 12$
Examiner tip: Expand two brackets first and simplify before multiplying by the third – trying to do all three at once is where terms get lost.

6Grade 8–9HigherNon-calculator3 marks

$a$ and $b$ are positive integers.

$$(ax + b)^2 - 3x(x - 4) \equiv x^2 + 32x + 25$$

Find the value of $a$ and the value of $b$. [3]

for expanding correctly: $a^2x^2 + 2abx + b^2 - 3x^2 + 12x$M1
for equating coefficients to form at least one correct equation, e.g. $a^2 - 3 = 1$, $b^2 = 25$ or $2ab + 12 = 32$M1
$a = 2$ and $b = 5$A1
Answer: $a = 2$, $b = 5$
Examiner tip: Remember the $-3x(x - 4)$ gives $+12x$; check your values in the $x$-coefficient equation ($2 \times 2 \times 5 + 12 = 32$).

7Grade 6–7HigherNon-calculator2 marks

Factorise fully $x^4 - 81$ [2]

for $(x^2 + 9)(x^2 - 9)$M1
$(x^2 + 9)(x + 3)(x - 3)$A1
Answer: $(x^2 + 9)(x + 3)(x - 3)$
Examiner tip: 'Fully' is the clue – $x^2 - 9$ is itself a difference of two squares, but $x^2 + 9$ does not factorise.

8Grade 1–3FoundationNon-calculator3 marks

(a) Choose the correct word from the box to describe each of the following.

expression equation formula identity

(i) $4x - 7$

(ii) $3y + 1 = 10$

(iii) $A = \frac{1}{2}bh$

(iv) $5(x - 2) \equiv 5x - 10$ [2]

(b) Simplify $p \times 4 \times p \times p$ [1]

(a) at least two of the four correct: (i) expression, (ii) equation, (iii) formula, (iv) identityB1
(a) all four correctB1
(b) $4p^3$ cao; do not accept $4ppp$ or $p^3 \times 4$ left unsimplifiedB1
Answer: (a) (i) expression (ii) equation (iii) formula (iv) identity (b) $4p^3$
Examiner tip: An identity uses $\equiv$ and is true for every value of $x$; an equation is only true for particular values – students often swap these two.

9Grade 4–5Non-calculator4 marks

The diagram shows a rectangle. All lengths are in centimetres.

(a) Show that the area of the rectangle, in cm², can be written as $x^2 + 5x - 6$ [2]

(b) Write an expression, in terms of $x$, for the perimeter of the rectangle. Give your answer in its simplest form. [2]

(a) $(x + 6)(x - 1)$ with at least three of the four terms correct, e.g. $x^2 - x + 6x - 6$M1
(a) fully correct expansion with $-x + 6x$ collected to reach $x^2 + 5x - 6$ with no errors (this is a show-that: the four-term line must be seen)A1
(b) $x + 6 + x - 1 + x + 6 + x - 1$ or $2(x + 6) + 2(x - 1)$ or $2x + 5$ seen as half the perimeterM1
(b) $4x + 10$; accept $2(2x + 5)$; ignore unitsA1
Answer: (a) $(x+6)(x-1) = x^2 - x + 6x - 6 = x^2 + 5x - 6$ (b) $4x + 10$
Examiner tip: For a show-that you must write out all four terms before collecting – jumping straight to the given answer scores M0 A0.

10Grade 4–5Non-calculator5 marks

(a) Sam writes

$$(x - 6)^2 = x^2 + 36$$

(i) Explain what Sam has done wrong. [1]

(ii) Expand and simplify $(x - 6)^2$ correctly. [2]

(b) Priya is asked to factorise fully $8x^2 + 12x$. She writes $2x(4x + 6)$.

Explain why Priya's answer is not fully factorised and write down the correct answer. [2]

(a)(i) e.g. Sam has squared each term separately / has not multiplied $(x - 6)(x - 6)$ / has missed out the $-12x$ term (the two $-6x$ terms); accept a correct counter-example, e.g. $x = 1$: $(1 - 6)^2 = 25$ but $1 + 36 = 37$B1
(a)(ii) $(x - 6)(x - 6)$ expanded to four terms with at least three correct, e.g. $x^2 - 6x - 6x + 36$M1
(a)(ii) $x^2 - 12x + 36$ caoA1
(b) e.g. $4x + 6$ still has a common factor of 2 / the HCF of $8x^2$ and $12x$ is $4x$, not $2x$B1
(b) $4x(2x + 3)$B1
Answer: (a)(i) She squared each term separately and missed the $-12x$ term. (ii) $x^2 - 12x + 36$ (b) $4x + 6$ still has a common factor 2; $4x(2x + 3)$
Examiner tip: Always write a squared bracket as two brackets multiplied – $(x-6)^2$ has a middle term, and $(-6)^2$ is $+36$.

11Grade 4–5Non-calculator4 marks

(a) Simplify $(3a^2b)^3$ [2]

(b) Simplify fully $\dfrac{20x^5y^4}{4x^2y}$ [2]

(a) two of $27$, $a^6$, $b^3$ correct in a product, e.g. $9a^6b^3$ or $27a^5b^3$B1
(a) $27a^6b^3$ caoB1
(b) two of $5$, $x^3$, $y^3$ correct in a product, e.g. $5x^3y^4$ or $16x^3y^3$B1
(b) $5x^3y^3$ caoB1
Answer: (a) $27a^6b^3$ (b) $5x^3y^3$
Examiner tip: The power outside the bracket applies to the number too: $3^3 = 27$, not 9 – and $y$ on its own means $y^1$, so $y^4 \div y = y^3$, not $y^4$.

12Grade 6–7HigherNon-calculator5 marks

(a) Factorise fully $3x^2 - 75$ [2]

(b) Factorise fully $4x^2 - 2x - 30$ [3]

(a) $3(x^2 - 25)$ or a correct partial factorisation such as $(3x - 15)(x + 5)$ or $(x - 5)(3x + 15)$M1
(a) $3(x - 5)(x + 5)$ (factors in any order)A1
(b) $2(2x^2 - x - 15)$, or a correct partial factorisation such as $(4x + 10)(x - 3)$ or $(2x + 5)(2x - 6)$M1
(b) $(2x \pm 5)(x \pm 3)$, i.e. factors of $2x^2 - x - 15$ correct apart from signs (with or without the factor 2); a correct but not fully factorised answer such as $(4x + 10)(x - 3)$ or $(2x + 5)(2x - 6)$ scores M1 M1 A0M1
(b) $2(2x + 5)(x - 3)$ (factors in any order); a fully correct answer scores 3A1
Answer: (a) $3(x - 5)(x + 5)$ (b) $2(2x + 5)(x - 3)$
Examiner tip: 'Fully' means take out the common number factor first – $(4x + 10)(x - 3)$ is correct but not fully factorised, so it loses the final mark.

13Grade 6–7HigherNon-calculator4 marks

(a) Expand and simplify $(3 + 2\sqrt{5})(4 - \sqrt{5})$ [2]

(b) $(2\sqrt{3} - 1)^2$ can be written in the form $a - b\sqrt{3}$, where $a$ and $b$ are integers.

Find the value of $a$ and the value of $b$. [2]

(a) four terms with at least three correct, e.g. $12 - 3\sqrt{5} + 8\sqrt{5} - 10$ (accept $-2\sqrt{25}$ or $-2 \times 5$ for $-10$)M1
(a) $2 + 5\sqrt{5}$A1
(b) $(2\sqrt{3} - 1)(2\sqrt{3} - 1)$ expanded to four terms with at least three correct, e.g. $12 - 2\sqrt{3} - 2\sqrt{3} + 1$M1
(b) $a = 13$, $b = 4$; accept $13 - 4\sqrt{3}$A1
Answer: (a) $2 + 5\sqrt{5}$ (b) $a = 13$, $b = 4$
Examiner tip: $2\sqrt{3} \times 2\sqrt{3}$ is $4 \times 3 = 12$, not $4\sqrt{9}$ left unsimplified or $2 \times 3 = 6$ – multiply the numbers and the surds separately.

14Grade 6–7HigherNon-calculator4 marks

(a) Without using a calculator, work out the value of $1003^2 - 997^2$

You must show your working. [2]

(b) Factorise $(x + 3)^2 - 5(x + 3)$ [2]

(a) $(1003 + 997)(1003 - 997)$ or $2000 \times 6$; working showing $1006009 - 994009$ also scores M1M1
(a) $12000$ cao (with working)A1
(b) $(x + 3)[(x + 3) - 5]$, or expansion to $x^2 + x - 6$M1
(b) $(x + 3)(x - 2)$ (either order)A1
Answer: (a) $12000$ (b) $(x + 3)(x - 2)$
Examiner tip: Spot the structure: $a^2 - b^2 = (a + b)(a - b)$ turns a horrible calculation into $2000 \times 6$, and $(x + 3)$ can be taken out as a common factor just like a single letter.

15Grade 8–9HigherNon-calculator3 marks

Factorise fully $3(2x - 1)^2 - 12$ [3]

$3[(2x - 1)^2 - 4]$, or correct expansion to $12x^2 - 12x - 9$M1
$3(2x - 1 - 2)(2x - 1 + 2)$, or $3(4x^2 - 4x - 3)$, or $(6x - 9)(2x + 1)$ or $(2x - 3)(6x + 3)$M1
$3(2x - 3)(2x + 1)$ (factors in any order)A1
Answer: $3(2x - 3)(2x + 1)$
Examiner tip: Treat $(2x - 1)$ as a single 'block': after taking out 3 you have (block)$^2 - 2^2$, a difference of two squares – forgetting the common factor 3 loses the final mark.

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