IGCSE Maths 0580 Algebraic Manipulation Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Algebra and Graphs · 2.2

Every algebraic manipulation question from 24 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.2. 30 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.

0580 Algebraic Manipulation past paper questions: Core (Papers 1 and 3)

12 marksCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q21(b)

Expand the brackets and simplify (x + 8)(x − 5).

Show solution
  1. Multiply each term in the first bracket by each term in the second bracket.
  2. x × x = x², x × (−5) = −5x, 8 × x = 8x, 8 × (−5) = −40.
  3. Collect like terms: −5x + 8x = 3x.
  4. (x + 8)(x − 5) = x² + 3x − 40
Answer x² + 3x − 40
24 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q18

(a) Simplify 3x − 2y + 5x + 7y. (b) Expand the brackets and simplify (3x + 5)(x − 7).

Show solution
Part (a)
  1. Collect the x terms: 3x + 5x = 8x. Collect the y terms: −2y + 7y = 5y.
  2. 8x + 5y
Answer 8x + 5y
Part (b)
  1. Multiply each term in the first bracket by each term in the second.
  2. 3x × x = 3x², 3x × (−7) = −21x, 5 × x = 5x, 5 × (−7) = −35.
  3. Collect like terms: −21x + 5x = −16x.
  4. (3x + 5)(x − 7) = 3x² − 16x − 35
Answer 3x² − 16x − 35
32 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q12

Factorise 5c − 15c²d.

Show solution
  1. Find the highest common factor of 5c and 15c²d. Numbers: 5. Letters: c.
  2. HCF = 5c. Divide each term by it: 5c ÷ 5c = 1 and 15c²d ÷ 5c = 3cd.
  3. 5c − 15c²d = 5c(1 − 3cd)
  4. Tip: Do not forget the 1. 5c × 1 = 5c.
Answer 5c(1 − 3cd)
42 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q20(b)

Factorise 6x² + 12x.

Show solution
  1. Find the highest common factor of 6x² and 12x. Numbers: 6. Letters: x.
  2. HCF = 6x. Divide each term: 6x² ÷ 6x = x and 12x ÷ 6x = 2.
  3. 6x² + 12x = 6x(x + 2)
Answer 6x(x + 2)
51 markCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q12

Factorise x² + 7x.

Show solution
  1. Both terms have x as a factor: x² = x × x and 7x = 7 × x.
  2. Take x outside the bracket: x(x + 7)
  3. Check: x(x + 7) = x² + 7x ✓
Answer x(x + 7)
62 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q12

Simplify 9w + 5q − 7w + 3q.

Show solution
  1. Collect the w terms: 9w − 7w = 2w.
  2. Collect the q terms: 5q + 3q = 8q.
  3. 2w + 8q
Answer 2w + 8q
72 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q18

Factorise 9x³ − 6x.

Show solution
  1. Find the highest common factor of 9x³ and 6x.
  2. Numbers: the HCF of 9 and 6 is 3. Letters: both terms contain at least one x.
  3. HCF = 3x. Divide each term by it: 9x³ ÷ 3x = 3x² and 6x ÷ 3x = 2.
  4. 9x³ − 6x = 3x(3x² − 2)
Answer 3x(3x² − 2)
85 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q22

(a) Expand the brackets and simplify (2x + 5)(x + 3).

(b) The diagram shows a rectangle (2x + 5) by (x + 3) with a rectangle 2x by (x + 1) cut out of one corner. Find an expression for the area of this shape. Give your answer in its simplest form.

0580/12 February/March 2026 Q22(b) – shape with sides 2x + 5, x + 3, 2x and x + 12x + 5x + 32xx + 1NOT TOSCALE
Show solution
Part (a)
  1. Multiply each term in the first bracket by each term in the second.
  2. 2x × x = 2x², 2x × 3 = 6x, 5 × x = 5x, 5 × 3 = 15.
  3. Collect like terms: 6x + 5x = 11x.
  4. (2x + 5)(x + 3) = 2x² + 11x + 15
Answer 2x² + 11x + 15
Part (b)
  1. Area of the shape = big rectangle − cut-out rectangle.
  2. Big rectangle = (2x + 5)(x + 3) = 2x² + 11x + 15 (from part (a)).
  3. Cut-out = 2x(x + 1) = 2x² + 2x.
  4. Area = 2x² + 11x + 15 − (2x² + 2x) = 2x² + 11x + 15 − 2x² − 2x.
  5. Area = 9x + 15
Answer 9x + 15
92 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q25

Expand and simplify (x + 3)(x − 2).

Show solution
  1. Multiply each term in the first bracket by each term in the second.
  2. x × x = x², x × (−2) = −2x, 3 × x = 3x, 3 × (−2) = −6.
  3. Collect like terms: −2x + 3x = x.
  4. (x + 3)(x − 2) = x² + x − 6
Answer x² + x − 6
104 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q25

(a) Expand and simplify (x − 5)(x + 2). (b) Factorise 5x³ + 10x.

Show solution
Part (a)
  1. Multiply each term in the first bracket by each term in the second.
  2. x × x = x², x × 2 = 2x, −5 × x = −5x, −5 × 2 = −10.
  3. Collect like terms: 2x − 5x = −3x.
  4. (x − 5)(x + 2) = x² − 3x − 10
Answer x² − 3x − 10
Part (b)
  1. Find the highest common factor of 5x³ and 10x. Numbers: 5. Letters: x.
  2. HCF = 5x. Divide each term: 5x³ ÷ 5x = x² and 10x ÷ 5x = 2.
  3. 5x³ + 10x = 5x(x² + 2)
Answer 5x(x² + 2)
115 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q17

(a) kˣ × k⁵ = k²⁰. Find the value of x. (b) Expand and simplify (x + 5)(x − 4). (c) Factorise 21r³ − 7r.

Show solution
Part (a)
  1. When multiplying powers of k, add the indices: kˣ × k⁵ = kx + 5.
  2. x + 5 = 20, so x = 15
Answer x = 15
Part (b)
  1. Multiply each term in the first bracket by each term in the second.
  2. x × x = x², x × (−4) = −4x, 5 × x = 5x, 5 × (−4) = −20.
  3. Collect like terms: −4x + 5x = x.
  4. (x + 5)(x − 4) = x² + x − 20
Answer x² + x − 20
Part (c)
  1. Find the highest common factor of 21r³ and 7r. Numbers: 7. Letters: r.
  2. HCF = 7r. Divide each term: 21r³ ÷ 7r = 3r² and 7r ÷ 7r = 1.
  3. 21r³ − 7r = 7r(3r² − 1)
Answer 7r(3r² − 1)
122 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q17

Factorise 21x − 7xy.

Show solution
  1. Find the highest common factor of 21x and 7xy. Numbers: 7. Letters: x.
  2. HCF = 7x. Divide each term: 21x ÷ 7x = 3 and 7xy ÷ 7x = y.
  3. 21x − 7xy = 7x(3 − y)
Answer 7x(3 − y)
132 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q19

Expand and simplify 3(5x + 2) − 4(x + 1).

Show solution
  1. Expand the first bracket: 3 × 5x + 3 × 2 = 15x + 6.
  2. Expand the second bracket. Take care with the minus: −4 × x − 4 × 1 = −4x − 4.
  3. Put them together: 15x + 6 − 4x − 4.
  4. Collect like terms: 15x − 4x = 11x and 6 − 4 = 2.
  5. 11x + 2
Answer 11x + 2
145 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q16

(a) Simplify. 6a + 4b − a − 5b

(b) Factorise. (i) 6x + 15y (ii) x²y − 5xy

Show solution
Part (a)
  1. Collect the a terms: 6a − a = 5a.
  2. Collect the b terms: 4b − 5b = −b.
  3. 6a + 4b − a − 5b = 5a − b
Answer 5a − b
Part (b)(i)
  1. Find the highest common factor of 6x and 15y: the numbers 6 and 15 share 3.
  2. 6x ÷ 3 = 2x and 15y ÷ 3 = 5y.
  3. 6x + 15y = 3(2x + 5y)
Answer 3(2x + 5y)
Part (b)(ii)
  1. Both terms contain x and y, so take out xy.
  2. x²y ÷ xy = x and 5xy ÷ xy = 5.
  3. x²y − 5xy = xy(x − 5)
  4. Check: xy × x − xy × 5 = x²y − 5xy ✓
Answer xy(x − 5)
154 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q23

(a) Factorise. 9x − 6xy

(b) Expand and simplify. (2x + 3)(x − 4)

Show solution
Part (a)
  1. Find the highest common factor of 9x and 6xy. Numbers: 3. Letters: x.
  2. HCF = 3x. Divide each term: 9x ÷ 3x = 3 and 6xy ÷ 3x = 2y.
  3. 9x − 6xy = 3x(3 − 2y)
Answer 3x(3 − 2y)
Part (b)
  1. Multiply each term in the first bracket by each term in the second.
  2. 2x × x = 2x², 2x × (−4) = −8x, 3 × x = 3x, 3 × (−4) = −12.
  3. Collect like terms: −8x + 3x = −5x.
  4. (2x + 3)(x − 4) = 2x² − 5x − 12
Answer 2x² − 5x − 12

0580 Algebraic Manipulation past paper questions: Extended (Papers 2 and 4)

162 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q16

Expand and simplify (3x + a)(x + 4a).

Show solution
  1. Multiply each term in the first bracket by each term in the second.
  2. 3x × x = 3x², 3x × 4a = 12ax, a × x = ax, a × 4a = 4a².
  3. Collect like terms: 12ax + ax = 13ax.
  4. 3x² + 13ax + 4a²
Answer 3x² + 13ax + 4a²
172 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q7

Factorise 6wx − 4xy.

Show solution
  1. Find the highest common factor of 6wx and 4xy. Numbers: 2. Letters: x is in both.
  2. HCF = 2x. Divide each term: 6wx ÷ 2x = 3w and 4xy ÷ 2x = 2y.
  3. 6wx − 4xy = 2x(3w − 2y)
Answer 2x(3w − 2y)
183 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q19

Expand and simplify (3x + 1)(x + 4)(x − 2).

Show solution
  1. Multiply the first two brackets: (3x + 1)(x + 4) = 3x² + 12x + x + 4 = 3x² + 13x + 4.
  2. Now multiply by (x − 2): (3x² + 13x + 4)(x − 2).
  3. 3x² × x = 3x³, 3x² × (−2) = −6x², 13x × x = 13x², 13x × (−2) = −26x, 4 × x = 4x, 4 × (−2) = −8.
  4. Collect like terms: −6x² + 13x² = 7x² and −26x + 4x = −22x.
  5. 3x³ + 7x² − 22x − 8
Answer 3x³ + 7x² − 22x − 8
193 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q21

(a) Write x² + 8x − 7 in the form (x + a)² + b, where a and b are integers.

(b) Write down the coordinates of the turning point of the curve y = x² + 8x − 7.

Show solution
Part (a)
  1. Halve the number in front of x: 8 ÷ 2 = 4, so start with (x + 4)².
  2. (x + 4)² = x² + 8x + 16, which is 16 too big compared with x² + 8x.
  3. So x² + 8x = (x + 4)² − 16, and x² + 8x − 7 = (x + 4)² − 16 − 7.
  4. (x + 4)² − 23
Answer (x + 4)² − 23
Part (b)
  1. (x + 4)² is never negative, so y is smallest when (x + 4)² = 0, that is, when x = −4.
  2. Then y = 0 − 23 = −23.
  3. Turning point = (−4, −23)
Answer (−4, −23)
204 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q26

(a) Write 3x² − 6x + 5 in the form a(x + b)² + c.

(b) Write down the coordinates of the turning point of the graph of y = 3x² − 6x + 5.

Show solution
Part (a)
  1. Take out 3 from the x² and x terms: 3x² − 6x + 5 = 3(x² − 2x) + 5.
  2. Complete the square inside: x² − 2x = (x − 1)² − 1.
  3. So 3[(x − 1)² − 1] + 5 = 3(x − 1)² − 3 + 5.
  4. 3(x − 1)² + 2
Answer 3(x − 1)² + 2
Part (b)
  1. (x − 1)² is never negative, so y is smallest when (x − 1)² = 0, that is, when x = 1.
  2. Then y = 0 + 2 = 2.
  3. Turning point = (1, 2)
Answer (1, 2)
211 markExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q4

Simplify a × 2a + 3a.

Show solution
  1. Multiply first: a × 2a = 2 × a × a = 2a².
  2. 2a² and 3a are not like terms (one has a², the other has a), so they cannot be added.
  3. 2a² + 3a
Answer 2a² + 3a
223 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q15

Factorise. (a) x² − 64 (b) 5x(x − 2y) + 6(x − 2y)²

Show solution
Part (a)
  1. x² − 64 is a difference of two squares: x² − 8².
  2. a² − b² = (a + b)(a − b), so x² − 64 = (x + 8)(x − 8)
Answer (x + 8)(x − 8)
Part (b)
  1. Both terms contain the bracket (x − 2y), so take it out as a common factor.
  2. 5x(x − 2y) + 6(x − 2y)² = (x − 2y)[5x + 6(x − 2y)]
  3. Simplify inside the square bracket: 5x + 6x − 12y = 11x − 12y.
  4. (x − 2y)(11x − 12y)
Answer (x − 2y)(11x − 12y)
233 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q22

A graph with equation y = x² + bx + c has a minimum point at (−5, 12). Find the value of b and the value of c.

Show solution
  1. A curve y = (x − p)² + q has its minimum point at (p, q).
  2. The minimum is at (−5, 12), so y = (x + 5)² + 12.
  3. Expand: (x + 5)² = x² + 10x + 25, so y = x² + 10x + 25 + 12 = x² + 10x + 37.
  4. Compare with y = x² + bx + c: b = 10, c = 37
Answer b = 10, c = 37
242 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q14

The minimum point on a quadratic curve is (−3, −5). (a) Find the equation of the line of symmetry of the curve. (b) Write the equation of the curve in the form y = (x + a)² + b.

Show solution
Part (a)
  1. A quadratic curve is symmetrical about the vertical line through its minimum point.
  2. The minimum is at x = −3, so the line of symmetry is x = −3
Answer x = −3
Part (b)
  1. A curve y = (x + a)² + b has its minimum at (−a, b).
  2. The minimum is (−3, −5), so a = 3 and b = −5: y = (x + 3)² − 5
Answer y = (x + 3)² − 5
254 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q15

Factorise. (a) x² − 7x + 12 (b) 5x + 10y + 6ny + 3nx

Show solution
Part (a)
  1. Find two numbers that multiply to +12 and add to −7: −3 and −4.
  2. x² − 7x + 12 = (x − 3)(x − 4)
Answer (x − 3)(x − 4)
Part (b)
  1. Rearrange into pairs that share a factor: (5x + 10y) + (3nx + 6ny).
  2. Factorise each pair: 5(x + 2y) + 3n(x + 2y).
  3. Both parts contain (x + 2y), so take it out: (x + 2y)(5 + 3n)
Answer (x + 2y)(5 + 3n)
261 markExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q6

Expand g(3 − 2g).

Show solution
  1. Multiply each term inside the bracket by g.
  2. g × 3 = 3g
  3. g × (−2g) = −2g²
  4. g(3 − 2g) = 3g − 2g²
Answer 3g − 2g²
273 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q17

Expand and simplify (x − 2)(2x + 3)(x + 4).

Show solution
  1. Multiply the first two brackets: (x − 2)(2x + 3) = 2x² + 3x − 4x − 6 = 2x² − x − 6.
  2. Multiply the result by (x + 4): every term times x, then every term times 4.
  3. 2x² × x − x × x − 6 × x = 2x³ − x² − 6x.
  4. 2x² × 4 − x × 4 − 6 × 4 = 8x² − 4x − 24.
  5. Collect like terms: 2x³ + (−x² + 8x²) + (−6x − 4x) − 24.
  6. 2x³ + 7x² − 10x − 24
Answer 2x³ + 7x² − 10x − 24
284 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q16

Expand and simplify. (a) 7(x + 2) + 4(3x − 5) (b) (3x − y)(5x + 2y)

Show solution
Part (a)
  1. Expand each bracket: 7(x + 2) = 7x + 14 and 4(3x − 5) = 12x − 20.
  2. Add them: 7x + 14 + 12x − 20.
  3. Collect like terms: 7x + 12x = 19x and 14 − 20 = −6.
  4. 19x − 6
Answer 19x − 6
Part (b)
  1. Multiply every term in the first bracket by every term in the second.
  2. 3x × 5x = 15x², 3x × 2y = 6xy, −y × 5x = −5xy, −y × 2y = −2y².
  3. Collect the xy terms: 6xy − 5xy = xy.
  4. 15x² + xy − 2y²
Answer 15x² + xy − 2y²
292 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q1

Simplify. 7c − 5d + c + 3d

Show solution
  1. Collect the c terms: 7c + c = 8c.
  2. Collect the d terms: −5d + 3d = −2d.
  3. 7c − 5d + c + 3d = 8c − 2d
Answer 8c − 2d
303 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q16

Expand and simplify. (x + 4)(x − 3)(3x + 2)

Show solution
  1. Multiply the first two brackets: (x + 4)(x − 3) = x² − 3x + 4x − 12 = x² + x − 12.
  2. Now multiply by (3x + 2): (x² + x − 12)(3x + 2).
  3. x² × 3x = 3x³, x² × 2 = 2x², x × 3x = 3x², x × 2 = 2x, −12 × 3x = −36x, −12 × 2 = −24.
  4. Collect like terms: 2x² + 3x² = 5x² and 2x − 36x = −34x.
  5. 3x³ + 5x² − 34x − 24
Answer 3x³ + 5x² − 34x − 24

Question papers used on this page

All algebraic manipulation questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/11/M/J/260580_s26_qp_11May/June 2026 · Paper 1 (Core)Q21(b)
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q18
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q12
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q16
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q7, Q19, Q21
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q26
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q20(b)
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q12
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q12
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q18, Q22
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q4
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q25
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q25
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q17
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q15, Q22
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q14, Q15
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q17
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q19
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q6
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q17
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q16
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q16
0580/12/M/J/250580_s25_qp_12May/June 2025 · Paper 1 (Core)Q23
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q1, Q16

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