Elite Mathematics / Linear / Topics / Factorisation IGCSE · 4MA1 · Topic Factorisation: past paper questions & worked solutions
Every Factorisation question from the 4MA1 papers in our bank, 2021–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
23 questions
117 marks in total
41% of papers (22 of 54) include it
1/7/15 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
1. Jun 2026 P1HR · Q8 Easy 2 marksShow worked solution Solution | Step 1 \[(-4)(-8)=32,\qquad -4+(-8)=-12.\]
Choose two numbers with product 32 and sum -12.
Solution | Step 2 \[x^2-12x+32=(x-4)(x-8).\]
The chosen numbers give the two constant terms in the factors.
Final answer \((x-4)(x-8)\)
2. Nov 2025 P2H · Q7 Medium 3 marksShow worked solution Factorise expression \[
x^2+2x-48=(x+8)(x-6)
\]
Factorise expression So \[ (x+8)(x-6)=0 \]
Final answer \(x=-8\) or \(x=6\)
3. Jun 2024 P2H · Q11 Medium 3 marksShow worked solution Factorise expression \[
x^2+9x-22=(x+11)(x-2)
\]
Factorise expression So \[ (x+11)(x-2)=0 \]
Final answer \(x=-11\) or \(x=2\)
4. May 2023 P1HR · Q7 Medium 3 marksShow worked solution (a) \[
4x(x-5)=4x^2-20x
\]
(b) \[
y^2-9y+20=(y-4)(y-5)
\]
Final answer (a) \(4x^2-20x\), (b) \((y-4)(y-5)\)
5. May 2023 P1H · Q14 Medium 3 marksShow worked solution Factorise expression \[
50g^2-18=2(25g^2-9)
\]
\[
=2(5g-3)(5g+3)
\]
Final answer \(2(5g-3)(5g+3)\)
6. Jun 2022 P2H · Q11 Medium 3 marksShow worked solution Factorise expression \[
x^2+5x-24=(x+8)(x-3)
\]
Factorise expression So \[ (x+8)(x-3)=0 \]
Final answer \(x=-8\) or \(x=3\)
7. May 2021 P2H · Q9 Medium 3 marksShow worked solution Factorise expression \[
x^2+2x-24=(x+6)(x-4)
\]
Factorise expression So \[ (x+6)(x-4)=0 \]
Final answer \(x=-6\) or \(x=4\)
8. Nov 2021 P2H · Q8 Medium 4 marksShow worked solution (a) (a) \((3x^2y)^0=1\).
(b)(i) \[
x^2-5x-36=(x-9)(x+4)
\]
Factorise expression (b)(ii) \(x=9\) or \(x=-4\).
Final answer \(1\), \((x-9)(x+4)\), \(x=9\) or \(x=-4\)
9. May 2026 4WM1H · Q9 Hard 5 marksShow worked solution (a) Take out the HCF HCF of \(15g^4h\) and \(20gh^3\) is \(5gh\): \[15g^4h-20gh^3=5gh(3g^3-4h^2)\]
(b)(i) Two numbers: product −28, sum 3 \(7\) and \(-4\): \[x^2+3x-28=(x+7)(x-4)\]
(b)(ii) Each bracket = 0 \[x=-7\quad\text{or}\quad x=4\]
Final answer (a) \(5gh(3g^3-4h^2)\) (b)(i) \((x+7)(x-4)\) (ii) \(x=-7,\ 4\)
10. Jun 2025 P1HR · Q8 Hard 5 marksShow worked solution (a) (a) \(a^6\times a^{10}=a^{16}\).
(b) (b) \(c^{30}\div c^{12}=c^{18}\).
(c)(i) \[
y^2-10y+21=(y-3)(y-7)
\]
(c)(ii) \[
y=3 \quad \text{or} \quad y=7
\]
Final answer \(a^{16}\), \(c^{18}\), \((y-3)(y-7)\), \(y=3\) or \(y=7\)
11. Jun 2022 P1HR · Q5 Hard 5 marksShow worked solution (a) (a) \(3c^3(c+4)=3c^4+12c^3\).
(b)(i) \[
x^2+8x-9=(x+9)(x-1)
\]
(b)(ii) \[
(x+9)(x-1)=0
\]
Final answer (a) \(3c^4+12c^3\), (b)(i) \((x+9)(x-1)\), (b)(ii) \(x=-9\) or \(x=1\)
12. Jun 2022 P1H · Q11 Hard 5 marksShow worked solution (a) \[
9x^2-4y^2=(3x)^2-(2y)^2=(3x-2y)(3x+2y)
\]
(b) \[
\frac78-\frac{x+3}{4x}
=\frac{7x}{8x}-\frac{2(x+3)}{8x}
\]
\[
=\frac{7x-2x-6}{8x}=\frac{5x-6}{8x}
\]
Final answer (a) \((3x-2y)(3x+2y)\), (b) \(\frac{5x-6}{8x}\)
13. Jan 2022 P1HR · Q9 Hard 5 marksShow worked solution (a) (a) \(x^4\times x^5=x^9\).
(b) (b) \((4y^2)^3=64y^6\).
(c) \[
n^2-7n+12=(n-3)(n-4)
\]
Final answer (a) \(x^9\), (b) \(64y^6\), (c) \((n-3)(n-4)\)
14. May 2021 P1H · Q9 Hard 5 marksShow worked solution (a) \[
25a^4c^7d+45a^9c^3h=5a^4c^3(5c^4d+9a^5h)
\]
(b) \[
(2x+5)^2=(2x+3)(2x-1)
\]
\[
4x^2+20x+25=4x^2+4x-3
\]
\[
16x=-28
\]
\[
x=-\frac74
\]
Final answer (a) \(5a^4c^3(5c^4d+9a^5h)\), (b) \(x=-\frac74\)
15. Jan 2021 P1H · Q9 Hard 5 marksShow worked solution (a) \[
(2x^3y^5)^4=16x^{12}y^{20}
\]
(b)(i) \[
x^2+5x-36=(x+9)(x-4)
\]
Factorise expression (b)(ii) \(x=-9\) or \(x=4\).
Final answer (a) \(16x^{12}y^{20}\), (b)(i) \((x+9)(x-4)\), (b)(ii) \(x=-9\) or \(x=4\)
16. Jun 2025 P1H · Q6 Hard 6 marksShow worked solution (a) \[
x-4=\frac{3+2x}{6}
\]
\[
6x-24=3+2x
\]
\[
4x=27
\]
\[
x=\frac{27}{4}
\]
(b)(i) \[
y^2-11y+30=(y-5)(y-6)
\]
(b)(ii) \[
y=5 \quad \text{or} \quad y=6
\]
Final answer \(x=\frac{27}{4}\), \((y-5)(y-6)\), \(y=5\) or \(y=6\)
17. May 2025 4WM1H · Q7 Hard 6 marksShow worked solution Factorise expression For part (a),
\[
x-4=\frac{3+2x}{6}
\]
Multiply by Multiply by \(6\).
\[
6x-24=3+2x
\]
\[
4x=27
\]
\[
x=\frac{27}{4}=6.75
\]
Factorise expression For part (b),
\[
y^2-11y+30=(y-5)(y-6)
\]
Factorise expression Hence,
\[
(y-5)(y-6)=0
\]
\[
y=5\quad\text{or}\quad y=6
\]
Final answer \(x=6.75\), \((y-5)(y-6)\), and \(y=5\) or \(6\).
18. Jun 2026 P2H · Q9 Hard 7 marksShow worked solution Part (a) | Step 1 Since \(n>0\), \(n^0=1\). Hence \(7n^0=7(1)=7\).
A nonzero number raised to power zero equals 1; only n is raised to zero.
Part (b) | Step 1 \[(2m^5p^4)^3=2^3m^{5\times3}p^{4\times3}=8m^{15}p^{12}.\]
Cube the coefficient and multiply each exponent by 3.
Part (c) | Step 1 The highest common factor of \(18c^5w^2\) and \(30cw^3\) is \(6cw^2\).
Take the numerical HCF and the smallest power of each shared variable.
Part (c) | Step 2 \[18c^5w^2+30cw^3=6cw^2(3c^4+5w).\]
Divide each original term by the common factor.
Part (d) | Step 1 Two numbers with product \(-27\) and sum \(6\) are \(9\) and \(-3\).
For a monic quadratic, find a factor pair with the required sum.
Part (d) | Step 2 \[x^2+6x-27=(x+9)(x-3).\]
The cross terms are -3x+9x=6x.
Final answer (a) \(7\)
(b) \(8m^{15}p^{12}\)
(c) \(6cw^2(3c^4+5w)\)
(d) \((x+9)(x-3)\)
19. Nov 2025 P1H · Q10 Hard 7 marksShow worked solution (a) (a) \((6m)^0=1\).
(b) \[
5y+20<7y+1
\]
\[
19<2y
\]
\[
y>\frac{19}{2}
\]
(c) \[
15w^2x^5+25w^3x^2=5w^2x^2(3x^3+5w)
\]
(d) \[
\frac{36a^5c^7}{12a^2c^3}=3a^3c^4
\]
Final answer \(1\), \(y>\frac{19}{2}\), \(5w^2x^2(3x^3+5w)\), \(3a^3c^4\)
20. May 2025 4WM1HR · Q9 Hard 7 marksShow worked solution (a) \[
(5k^3n^4)^2=25k^6n^8
\]
(b) \[
12m^2y^4+18m^3y^3=6m^2y^3(2y+3m)
\]
(c) \[
x^2-13x+42=(x-6)(x-7)
\]
Calculate value So \[ x=6 \quad \text{or} \quad x=7 \]
Final answer \(25k^6n^8\), \(6m^2y^3(2y+3m)\), \((x-6)(x-7)\), \(x=6,7\).
21. May 2023 P1H · Q6 Hard 8 marksShow worked solution (a) \[
(2c^4d^7)^3=8c^{12}d^{21}
\]
Factorise expression (b) \(5y^0=5\), since \(y^0=1\).
(c) \[
16a^2b^3+20a^3b=4a^2b(4b^2+5a)
\]
(d)(i) \[
x^2+9x-22=(x+11)(x-2)
\]
Factorise expression (d)(ii) \(x=-11\) or \(x=2\).
Final answer \(8c^{12}d^{21}\), \(5\), \(4a^2b(4b^2+5a)\), \((x+11)(x-2)\), \(x=-11\) or \(x=2\)
22. May-Nov 2020 P1H · Q7 Hard 8 marksShow worked solution (a) \[
5(4-x)=7-3x
\]
\[
20-5x=7-3x
\]
\[
13=2x
\]
\[
x=\frac{13}{2}
\]
(b) \[
16m^3g^3+24m^2g^5=8m^2g^3(2m+3g^2)
\]
(c)(i) \[
y^2-2y-48=(y-8)(y+6)
\]
Factorise expression (c)(ii) \(y=8\) or \(y=-6\).
Final answer \(x=\frac{13}{2}\), \(8m^2g^3(2m+3g^2)\), \((y-8)(y+6)\), \(y=8\) or \(y=-6\)
23. Nov 2025 4WM1H · Q8 Hard 9 marksShow worked solution (a) \[
5w(3w-4w^2)=15w^2-20w^3
\]
(b) \[
(4a^3c^5)^3=64a^9c^{15}
\]
(c) \[
14m^3p^2+21m^2p^4=7m^2p^2(2m+3p^2)
\]
(d) \[
x^2-10x+24=(x-4)(x-6)
\]
Calculate value So \[ x=4 \quad \text{or} \quad x=6 \]
Final answer \(15w^2-20w^3\), \(64a^9c^{15}\), \(7m^2p^2(2m+3p^2)\), \((x-4)(x-6)\), \(x=4,6\).
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