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IGCSE · 4MA1 · Topic

Expanding Brackets: past paper questions & worked solutions

Every Expanding Brackets question from the 4MA1 papers in our bank, 2021–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.

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21questions
105marks in total
39%of papers (21 of 54) include it
0/8/13easy / medium / hard

Questions, easiest first

Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.

1. Jun 2025 P2H · Q12

Medium3 marks
Jun 2025 P2H Question 12 (3 marks): Expanding Brackets – 12 Expand and simplify 3 2 5 7 4 x x x (Total for Question 12 is 3 marks)
Show worked solution
  1. Expand brackets
    \[ 3x(2x+5)(7x-4)=3x(14x^2+27x-20) \]
    \[ =42x^3+81x^2-60x \]
Final answer

\(42x^3+81x^2-60x\)

2. Nov 2024 P1H · Q15

Medium3 marks
Nov 2024 P1H Question 15 (3 marks): Expanding Brackets – 15 Write 3 2 1 5 4 x x x in the form ax bx cx 3 2 + + where a, b and c are integers. (Total for Question 15 is 3 marks)
Show worked solution
  1. Expand brackets
    \[ 3x(2x-1)(5x+4) \]
  2. Expand brackets

    First expand the two brackets:

    \[ (2x-1)(5x+4)=10x^2+8x-5x-4 \]
    \[ =10x^2+3x-4 \]
  3. Expand brackets

    Now multiply by \(3x\):

    \[ 3x(10x^2+3x-4)=30x^3+9x^2-12x \]
Final answer

\(30x^3+9x^2-12x\)

3. Nov 2023 P2H · Q16

Medium3 marks
Nov 2023 P2H Question 16 (3 marks): Expanding Brackets – 16 Expand and simplify (2x + 3)(x - 5)(x + 4) (Total for Question 16 is 3 marks)
Show worked solution
  1. Expand brackets
    \[ (x-5)(x+4)=x^2-x-20 \]
    \[ (2x+3)(x^2-x-20)=2x^3+x^2-43x-60 \]
Final answer

\(2x^3+x^2-43x-60\)

4. Nov 2021 P1H · Q1

Medium3 marks
Nov 2021 P1H Question 1 (3 marks): Expanding Brackets – 1 (a) Simplify e e 8 2 ÷ (1) (b) Expand and simplify (x - 3)(x + 1) (2) (Total for Question 1 is 3 marks)
Show worked solution
  1. (a)

    (a) \(e^8\div e^2=e^6\).

  2. (b)

    (b) \((x-3)(x+1)=x^2-2x-3\).

Final answer

(a) \(e^6\), (b) \(x^2-2x-3\)

5. May 2021 P1H · Q13

Medium3 marks
May 2021 P1H Question 13 (3 marks): Expanding Brackets – 13 Expand and simplify 4x(3x + 1)(2x - 3) Show your working clearly. (Total for Question 13 is 3 marks)
Show worked solution
  1. Expand brackets
    \[ 4x(3x+1)(2x-3)=4x(6x^2-7x-3) \]
    \[ =24x^3-28x^2-12x \]
Final answer

\(24x^3-28x^2-12x\)

6. May-Nov 2020 P1HR · Q14

Medium3 marks
May-Nov 2020 P1HR Question 14 (3 marks): Expanding Brackets – 14 Expand and simplify (4x + 1)(x - 3)(5x + 6) (Total for Question 14 is 3 marks)
Show worked solution
  1. Expand brackets
    \[ (4x+1)(x-3)(5x+6) \]
    \[ (x-3)(5x+6)=5x^2-9x-18 \]
    \[ (4x+1)(5x^2-9x-18)=20x^3-31x^2-81x-18 \]
Final answer

\(20x^3-31x^2-81x-18\)

7. Jun 2023 P2HR · Q1

Medium4 marks
Jun 2023 P2HR Question 1 (4 marks): Expanding Brackets – 1 P = m2 - 4c (a) Work out the value of P when m = -5 and c = 3 P = (2) (b) Expand and simplify (x + 5)(x - 7) (2) (Total for Question 1 is 4 marks)
Show worked solution
  1. Expand brackets
    \[ P=m^2-4c=(-5)^2-4(3)=25-12=13 \]
    \[ (x+5)(x-7)=x^2-7x+5x-35=x^2-2x-35 \]
Final answer

\(P=13\), and \(x^2-2x-35\)

8. Jan 2022 P2H · Q1

Medium4 marks
Jan 2022 P2H Question 1 (4 marks): Expanding Brackets – 1 (a) Expand and simplify (y + 4)(2 - y) (2) (b) Factorise fully 15b5c - 35b3c9 (2) (Total for Question 1 is 4 marks)
Show worked solution
  1. (a)
    \[ (y+4)(2-y)=2y-y^2+8-4y=-y^2-2y+8 \]
  2. (b)
    \[ 15b^5c-35b^3c^9=5b^3c(3b^2-7c^8) \]
Final answer

(a) \(-y^2-2y+8\), (b) \(5b^3c(3b^2-7c^8)\)

9. Jun 2025 P2HR · Q13

Hard5 marks
Jun 2025 P2HR Question 13 (5 marks): Expanding Brackets – 13 (a) Factorise 4 25 2 2 x y - (2) (b) Show that 4 3 2 5 x x x ( )( ) can be written in the form ax bx cx 3 2 + + where a, b and c are integers to be found. (3) (Total for Question 13 is 5 marks)
Show worked solution
  1. (a)
    \[ 4x^2-25y^2=(2x-5y)(2x+5y) \]
  2. (b)
    \[ 4x(x+3)(2x-5) \]
  3. Factorise expression

    First, \[ (x+3)(2x-5)=2x^2+x-15 \]

  4. Factorise expression

    Then \[ 4x(2x^2+x-15)=8x^3+4x^2-60x \]

  5. Factorise expression

    So \(a=8\), \(b=4\), and \(c=-60\).

Final answer

(a) \((2x-5y)(2x+5y)\), (b) \(8x^3+4x^2-60x\)

10. Jan 2021 P1HR · Q16

Hard5 marks
Jan 2021 P1HR Question 16 (5 marks): Expanding Brackets – 16 (a) Expand and simplify (x + 4)(x - 2)(x + 1) (3) (b) Express x 2 - 10x + 40 in the form (x + a) 2 + b, where a and b are integers. (2) (Total for Question 16 is 5 marks)
Show worked solution
  1. (a)
    \[ (x+4)(x-2)(x+1) \]
    \[ (x-2)(x+1)=x^2-x-2 \]
    \[ (x+4)(x^2-x-2)=x^3+3x^2-6x-8 \]
  2. (b)
    \[ x^2-10x+40=(x-5)^2+15 \]
Final answer

(a) \(x^3+3x^2-6x-8\), (b) \((x-5)^2+15\)

11. Jun 2026 P1H · Q12

Hard6 marks
Jun 2026 P1H Question 12 (6 marks): Expanding Brackets – 12 2 8 64 3 x = (a) Find the value of x Show clear algebraic working. x = (3) (b) Expand and simplify 4 5 2 3 t t t ( )( ) � � (3) (Total for Question 12 is 6 marks)
Show worked solution
  1. Part (a) | Step 1
    \[\frac{2^{x/3}}{8^x}=\frac{2^{x/3}}{2^{3x}}=2^{x/3-3x}=2^{-8x/3}.\]

    Write 8 as a power of 2 and subtract exponents when dividing.

  2. Part (a) | Step 2
    \[2^{-8x/3}=64=2^6\implies-\frac{8x}{3}=6\implies x=-\frac94.\]

    Equal powers of base 2 have equal exponents.

  3. Part (b) | Step 1
    \[(t+5)(2t-3)=2t^2-3t+10t-15=2t^2+7t-15.\]

    Expand the two brackets and collect like terms.

  4. Part (b) | Step 2
    \[4t(t+5)(2t-3)=4t(2t^2+7t-15)=8t^3+28t^2-60t.\]

    Multiply every term by 4t.

Final answer

(a) \(x=-\frac94\)

(b) \(8t^3+28t^2-60t\)

12. Nov 2025 4WM1H · Q13

Hard6 marks
Nov 2025 4WM1H Question 13 (6 marks): Expanding Brackets – 13 (a) Expand and simplify 7x(3x + 2)(2x – 5) (3) (b) Solve 9 2 5 7 5 y Show clear algebraic working. y = (3) (Total for Question 13 is 6 marks)
Show worked solution
  1. (a)
    \[ 7x(3x+2)(2x-5) \]
    \[ =7x(6x^2-11x-10) \]
    \[ =42x^3-77x^2-70x \]
  2. (b)
    \[ \frac{9}{2y}+\frac57=5 \]
    \[ \frac{9}{2y}=\frac{30}{7} \]
    \[ 63=60y \]
    \[ y=\frac{21}{20} \]
Final answer

\(42x^3-77x^2-70x\), \(y=\frac{21}{20}\).

13. Nov 2025 P1H · Q14

Hard6 marks
Nov 2025 P1H Question 14 (6 marks): Expanding Brackets – 14 (a) Expand and simplify 7x(3x + 2)(2x - 5) (3) (b) Solve 9 2 5 7 5 y Show clear algebraic working. y = (3) (Total for Question 14 is 6 marks)
Show worked solution
  1. (a)
    \[ 7x(3x+2)(2x-5)=7x(6x^2-11x-10) \]
    \[ =42x^3-77x^2-70x \]
  2. (b)
    \[ \frac9{2y}+\frac57=5 \]
    \[ \frac9{2y}=\frac{30}{7} \]
    \[ 63=60y \]
Final answer

(a) \(42x^3-77x^2-70x\), (b) \(y=\frac{21}{20}\)

14. May 2025 4WM1H · Q11

Hard6 marks
May 2025 4WM1H Question 11 (6 marks): Expanding Brackets – 11 (a) Expand and simplify 5 5 4 2 3 x x x (3) (b) Express 9 4 7 5 y y as a single fraction in its simplest form. (3) (Total for Question 11 is 6 marks)
Show worked solution
  1. Expand brackets
    \[ 5x(5x+4)(2x-3) \]
  2. Expand brackets

    First,

    \[ (5x+4)(2x-3)=10x^2-15x+8x-12=10x^2-7x-12 \]
  3. Expand brackets

    Then multiply by \(5x\).

    \[ 5x(10x^2-7x-12)=50x^3-35x^2-60x \]
  4. Expand brackets

    For part (b), use denominator \(20y\).

    \[ \frac94+\frac{y-7}{5y}=\frac{45y}{20y}+\frac{4(y-7)}{20y} \]
    \[ =\frac{49y-28}{20y} \]
Final answer

\(50x^3-35x^2-60x\) and \(\frac{49y-28}{20y}\).

15. May 2025 4WM1HR · Q15

Hard6 marks
May 2025 4WM1HR Question 15 (6 marks): Expanding Brackets – 15 (a) Solve 3 4 18 0 2 x x Show your working clearly. Give your solutions correct to 3 significant figures. (3) (b) Expand and simplify 2 3 4 1 x x x (3) (Total for Question 15 is 6 marks)
Show worked solution
  1. Simplify surd

    (a) For \(3x^2+4x-18=0\),

    \[ x=\frac{-4\pm\sqrt{4^2-4(3)(-18)}}{2(3)} \]
    \[ x\approx 1.87 \quad \text{or} \quad x\approx -3.21 \]
  2. (b)
    \[ (2x-3)(x+4)(x-1) \]
    \[ =(2x^2+5x-12)(x-1) \]
    \[ =2x^3+3x^2-17x+12 \]
Final answer

\(x=1.87\) or \(x=-3.21\); \(2x^3+3x^2-17x+12\).

16. May 2024 P1HR · Q14

Hard6 marks
May 2024 P1HR Question 14 (6 marks): Expanding Brackets – 14 (a) Expand and simplify  (3) (b) Simplify fully a b a b 3 9 5 1 2     (3) (Total for Question 14 is 6 marks)
Show worked solution
  1. (a)
    \[ (2-x)(4+x)=8-2x-x^2 \]
    \[ (3x+1)(8-2x-x^2)=-3x^3-7x^2+22x+8 \]
  2. (b)
    \[ \left(\frac{a^3b}{a^9b^5}\right)^{-1/2} =\left(\frac1{a^6b^4}\right)^{-1/2}=a^3b^2 \]
Final answer

(a) \(-3x^3-7x^2+22x+8\), (b) \(a^3b^2\)

17. Jun 2022 P1HR · Q16

Hard6 marks
Jun 2022 P1HR Question 16 (6 marks): Expanding Brackets – 16 (a) Expand and simplify (3x - 1)(x + 2)(3x + 1) (3) (b) Simplify fully 2 8 5 2 2 x xy     - (3) (Total for Question 16 is 6 marks)
Show worked solution
  1. (a)
    \[ (3x-1)(x+2)(3x+1)=(9x^2-1)(x+2) \]
    \[ =9x^3+18x^2-x-2 \]
  2. (b)
    \[ \left(\frac{2x^5}{8xy^2}\right)^{-2} =\left(\frac{x^4}{4y^2}\right)^{-2} \]
    \[ =\left(\frac{4y^2}{x^4}\right)^2=\frac{16y^4}{x^8} \]
Final answer

(a) \(9x^3+18x^2-x-2\), (b) \(\frac{16y^4}{x^8}\)

18. Jan 2022 P1H · Q13

Hard6 marks
Jan 2022 P1H Question 13 (6 marks): Expanding Brackets – 13 (a) Expand and simplify 5x(x + 2)(3x - 4) (3) (b) Simplify completely 16 8 20 3 4 w y     - (3) (Total for Question 13 is 6 marks)
Show worked solution
  1. (a)
    \[ 5x(x+2)(3x-4)=5x(3x^2+2x-8) \]
    \[ =15x^3+10x^2-40x \]
  2. (b)
    \[ \left(\frac{16w^8}{y^{20}}\right)^{-3/4} =16^{-3/4}w^{-6}y^{15} \]
  3. Expand brackets

    Since \(16^{3/4}=8\), \[ =\frac{y^{15}}{8w^6} \]

Final answer

(a) \(15x^3+10x^2-40x\), (b) \(\frac{y^{15}}{8w^6}\)

19. Jan 2021 P1H · Q12

Hard6 marks
Jan 2021 P1H Question 12 (6 marks): Expanding Brackets – 12 (a) Express 4 2 3 1 x x - - + as a single fraction. Give your answer in its simplest form. (3) (b) Expand and simplify 2x(x - 5)(x - 3) (3) (Total for Question 12 is 6 marks)
Show worked solution
  1. (a)
    \[ \frac4{x-2}-\frac3{x+1} =\frac{4(x+1)-3(x-2)}{(x-2)(x+1)} \]
    \[ =\frac{4x+4-3x+6}{(x-2)(x+1)} =\frac{x+10}{(x-2)(x+1)} \]
  2. (b)
    \[ 2x(x-5)(x-3)=2x(x^2-8x+15) \]
    \[ =2x^3-16x^2+30x \]
Final answer

(a) \(\frac{x+10}{(x-2)(x+1)}\), (b) \(2x^3-16x^2+30x\)

20. Nov 2021 P2H · Q12

Hard7 marks
Nov 2021 P2H Question 12 (7 marks): Expanding Brackets – 12 (a) Simplify (64p9q12) 2 3 (2) (b) Write as a single fraction 2 3 4 5 9 10 x x x + - Give your answer in its simplest form. (2) (c) Expand and simplify 4x(x - 5)(2x + 3) Show your working clearly. (3) (Total for
Show worked solution
  1. (a)
    \[ (64p^9q^{12})^{2/3}=16p^6q^8 \]
  2. (b)
    \[ \frac{2}{3x}+\frac{4}{5x}-\frac{9}{10x} =\frac{20+24-27}{30x}=\frac{17}{30x} \]
  3. (c)
    \[ 4x(x-5)(2x+3)=4x(2x^2-7x-15) \]
    \[ =8x^3-28x^2-60x \]
Final answer

(a) \(16p^6q^8\), (b) \(\frac{17}{30x}\), (c) \(8x^3-28x^2-60x\)

21. Jan 2021 P2HR · Q1

Hard8 marks
Jan 2021 P2HR Question 1 (8 marks): Expanding Brackets – 1 w = 5y 2 - y 3 (a) Work out the value of w when y = -2 w = (2) (b) Factorise fully 8p 2 - 2p (2) (c) Expand 4t(3t - 2) (2) (d) Expand and simplify (5x - 2)(x + 4) (2) (Total for Question 1 is 8 marks)
Show worked solution
  1. Expand brackets

    For \(w=5y^2-y^3\) and \(y=-2\): \[ w=5(-2)^2-(-2)^3=20+8=28 \]

    \[ 8p^2-2p=2p(4p-1) \]
    \[ 4t(3t-2)=12t^2-8t \]
    \[ (5x-2)(x+4)=5x^2+18x-8 \]
Final answer

\(28\), \(2p(4p-1)\), \(12t^2-8t\), \(5x^2+18x-8\)

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