Algebraic manipulation: simplifying, expanding & factorising – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

(a) Simplify $5p + 3q - 2p + q$ [2]

(b) Simplify $m \times m \times m$ [1]

(a) one correct term, $3p$ or $4q$, in an expression of two termsB1
(a) $3p + 4q$ (accept $4q + 3p$)B1
(b) $m^3$ (do not accept $3m$)B1
(c) $12xy$ (accept $12yx$; do not accept $12x \times y$ as final answer)B1
Answer: (a) $3p + 4q$ (b) $m^3$ (c) $12xy$
Examiner tip: The lone $q$ means $1q$, so $3q + q = 4q$, not $3q$; and $m \times m \times m$ is $m^3$, not $3m$.

2Grade 1–3Calculator4 marks

(a) Work out the value of $3a^2 - 2b$ when $a = -4$ and $b = 5$ [2]

(b) Expand $4(3x - 7)$ [1]

(a) for correct substitution, e.g. $3 \times (-4)^2 - 2 \times 5$ or $3 \times 16 - 10$ or $48$ seenM1
(a) $38$ caoA1
(b) $12x - 28$B1
(c) $5(3y + 2)$B1
Answer: (a) $38$ (b) $12x - 28$ (c) $5(3y+2)$
Examiner tip: Square $-4$ before multiplying by 3: $(-4)^2 = 16$, so $3a^2 = 48$, not $-48$ and not $144$.

3Grade 4–5Calculator5 marks

(a) Expand and simplify $3(2x + 5) - 4(x - 2)$ [2]

(b) Factorise fully $12x^2y + 18xy^2$ [2]

(a) for $6x + 15$ and $-4x + 8$ seen (allow one sign or arithmetic error in the four terms)M1
(a) $2x + 23$ caoA1
(b) for a correct partial factorisation with at least two of $6$, $x$, $y$ taken out, e.g. $6x(2xy + 3y^2)$, $3xy(4x + 6y)$ or $2xy(6x + 9y)$; or $6xy(\ldots)$ with at most one error in the bracketM1
(b) $6xy(2x + 3y)$A1
(c) $2t^3 - 10t$B1
Answer: (a) $2x + 23$ (b) $6xy(2x + 3y)$ (c) $2t^3 - 10t$
Examiner tip: In (a) $-4 \times -2 = +8$; writing $-8$ gives $2x + 7$ and loses the accuracy mark. In (b) 'fully' means the HCF $6xy$ must be outside.

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4Grade 4–5Calculator5 marks

(a) Expand and simplify $(x + 9)(x - 4)$ [2]

(b) Factorise $x^2 - 3x - 40$ [2]

Show that Sam is wrong. [1]

(a) for 3 of the 4 terms correct in $x^2 - 4x + 9x - 36$M1
(a) $x^2 + 5x - 36$A1
(b) for $(x \pm 8)(x \pm 5)$M1
(b) $(x - 8)(x + 5)$ (either order)A1
(c) for a correct expansion $x^2 + 10x + 25$ (or $x^2 + 5x + 5x + 25$) with statement that it is not $x^2 + 25$; or a correct counter-example, e.g. $x = 1$: $(1+5)^2 = 36$ but $1 + 25 = 26$C1
Answer: (a) $x^2 + 5x - 36$ (b) $(x - 8)(x + 5)$ (c) $(x+5)^2 = x^2 + 10x + 25 \neq x^2 + 25$
Examiner tip: $(x+5)^2$ means $(x+5)(x+5)$: four terms before simplifying, so the middle term $10x$ never disappears.

5Grade 4–5Calculator4 marks

The diagram shows a rectangle of length $(2x + 3)$ cm and width $(x + 4)$ cm.

A square of side $x$ cm is removed from one corner. The remaining region is shaded.

(a) Show that the area, in cm$^2$, of the shaded region is $x^2 + 11x + 12$ [3]

(b) Find the area of the shaded region when $x = 6$ [1]

(a) for $(2x + 3)(x + 4) - x^2$ seen (rectangle minus square), or for splitting the shaded region into rectangles, e.g. $x(x + 3) + 4(2x + 3)$ oeM1
(a) for a correct expansion with at least 3 of the 4 terms correct in $2x^2 + 8x + 3x + 12$ (or, for the split method, $x^2 + 3x$ and $8x + 12$)M1
(a) for $2x^2 + 11x + 12 - x^2 = x^2 + 11x + 12$ (or $x^2 + 3x + 8x + 12 = x^2 + 11x + 12$) with no errors seen (answer given, so all steps must be shown)A1
(b) $114$ (cm$^2$) (allow from substituting into the rectangle minus square: $15 \times 10 - 36$)B1
Answer: (a) $(2x+3)(x+4) - x^2 = 2x^2 + 11x + 12 - x^2 = x^2 + 11x + 12$ (b) $114$ cm$^2$
Examiner tip: This is a 'show that': the final line is given, so you must show the full expansion $2x^2 + 8x + 3x + 12$ and the subtraction of $x^2$.

6Grade 6–7HigherCalculator5 marks

Factorise fully

(a) $2x^2 + 7x - 15$ [2]

(b) $9y^2 - 49$ [1]

(a) for $(2x \pm 3)(x \pm 5)$, or for $2x^2 + 10x - 3x - 15$ followed by a correct partial factorisationM1
(a) $(2x - 3)(x + 5)$A1
(b) $(3y + 7)(3y - 7)$B1
(c) for taking out the common factor, $2(25 - m^2)$, or for $(10 - 2m)(5 + m)$ or $(5 - m)(10 + 2m)$M1
(c) $2(5 - m)(5 + m)$ oe, e.g. $-2(m - 5)(m + 5)$A1
Answer: (a) $(2x - 3)(x + 5)$ (b) $(3y + 7)(3y - 7)$ (c) $2(5 - m)(5 + m)$
Examiner tip: In (c), take out the common factor 2 first; $(10 - 2m)(5 + m)$ is not fully factorised and only scores M1.

7Grade 6–7HigherCalculator5 marks

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

(b) $(x + a)(x - 6) \equiv x^2 + bx - 18$ where $a$ and $b$ are integers.

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

(a) for correct expansion of any two of the brackets with at least 3 of 4 terms correct, e.g. $x^2 + x - 6$, $2x^2 + 7x + 3$ or $2x^2 - 3x - 2$M1
(a) (dep) for multiplying their quadratic by the third bracket with at least 4 of the 6 terms correct (ft their quadratic), e.g. $2x^3 + x^2 + 2x^2 + x - 12x - 6$M1
(a) $2x^3 + 3x^2 - 11x - 6$ caoA1
(b) for $-6a = -18$ or $a = 3$, or for $x^2 - 6x + ax - 6a$ seenM1
(b) $a = 3$ and $b = -3$A1
Answer: (a) $2x^3 + 3x^2 - 11x - 6$ (b) $a = 3$, $b = -3$
Examiner tip: Expand two brackets fully and simplify before multiplying by the third; most lost marks come from dropping one of the six terms.

8Grade 8–9HigherCalculator6 marks

(a) Write $x^2 - 10x + 31$ in the form $(x + p)^2 + q$ where $p$ and $q$ are integers. [2]

(b) Hence explain why $x^2 - 10x + 31 > 0$ for all values of $x$. [1]

(a) for $(x - 5)^2$ seen, or for $p = -5$M1
(a) $(x - 5)^2 + 6$A1
(b) for an explanation that $(x - 5)^2 \geq 0$ so the expression is at least $6$ (ft their $q > 0$)C1
(c) for $2(x^2 + 6x) + 5$ or $2\left(x^2 + 6x + \tfrac{5}{2}\right)$, or $a = 2$ statedM1
(c) for $2(x + 3)^2 - 18 + 5$ oe, or $2\left[(x + 3)^2 - 9 + \tfrac{5}{2}\right]$M1
(c) $2(x + 3)^2 - 13$A1
Answer: (a) $(x - 5)^2 + 6$ (b) $(x-5)^2 \geq 0$ so the expression $\geq 6 > 0$ (c) $2(x + 3)^2 - 13$
Examiner tip: In (c) the $-9$ from completing the square is inside the factor 2, so subtract $2 \times 9 = 18$, not 9; $2(x+3)^2 - 4$ is the common wrong answer.

9Grade 8–9HigherCalculator6 marks

(a) Simplify fully $\dfrac{2x^2 + 5x - 3}{x^2 - 9}$ [3]

(b) Write $\dfrac{3}{x + 2} - \dfrac{2}{x - 1}$ as a single fraction in its simplest form. [3]

(a) for $(2x - 1)(x + 3)$M1
(a) for $(x - 3)(x + 3)$M1
(a) $\dfrac{2x - 1}{x - 3}$A1
(b) for a correct common denominator, e.g. $(x + 2)(x - 1)$ or $x^2 + x - 2$, with at least one numerator correct, e.g. $\dfrac{3(x - 1) - \ldots}{(x + 2)(x - 1)}$M1
(b) for $\dfrac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)}$ or $\dfrac{3x - 3 - 2x - 4}{(x + 2)(x - 1)}$M1
(b) $\dfrac{x - 7}{(x + 2)(x - 1)}$ (accept $\dfrac{x - 7}{x^2 + x - 2}$)A1
Answer: (a) $\dfrac{2x - 1}{x - 3}$ (b) $\dfrac{x - 7}{(x + 2)(x - 1)}$
Examiner tip: In (b) the minus sign applies to the whole of $2(x + 2)$, giving $-2x - 4$; writing $-2x + 4$ gives $x + 1$ on top and loses the accuracy mark.

10Grade 6–7HigherCalculator3 marks

$n$ is an integer.

Prove algebraically that $(3n + 2)^2 - (n - 2)^2$ is always a multiple of 8. [3]

for a correct expansion of either bracket, $9n^2 + 12n + 4$ or $n^2 - 4n + 4$M1
for $8n^2 + 16n$ from correct working (or $(3n + 2 + n - 2)(3n + 2 - n + 2) = 4n(2n + 4)$ using the difference of two squares)M1
for $8n^2 + 16n = 8(n^2 + 2n)$ oe, e.g. $8n(n + 2)$, with a conclusion that this is a multiple of 8C1
Answer: $(3n+2)^2 - (n-2)^2 = 9n^2 + 12n + 4 - n^2 + 4n - 4 = 8n^2 + 16n = 8(n^2 + 2n)$, which is a multiple of 8.
Examiner tip: Put the second expansion in a bracket before subtracting: $-(n^2 - 4n + 4) = -n^2 + 4n - 4$; and finish by writing $8 \times$(an integer) with a statement.

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