Question Simulator

IGCSE Additional Mathematics (0606)

Factors of polynomials Banker

11 questions · about 5% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q3 (a, b, c)7 marksFactor and remainder theorems · Solving cubic equations

2025

  • 2025 Oct/Nov · Paper 12 · Q2 (a, b)7 marksFactor and remainder theorems · Solving cubic equations
  • 2025 May/June · Paper 12 · Q5 (a, b, c)10 marksFactor and remainder theorems · Solving cubic equations

2024

  • 2024 Oct/Nov · Paper 22 · Q3 (a, b)5 marksFactor and remainder theorems · Solving cubic equations
  • 2024 Oct/Nov · Paper 22 · Q10 (b)3 marksSolving cubic equations
  • 2024 May/June · Paper 22 · Q4 (a, b(i), b(ii))5 marksFactor and remainder theorems · Solving cubic equations

2023

  • 2023 Oct/Nov · Paper 12 · Q6 (a, b)10 marksFactor and remainder theorems · Solving cubic equations
  • 2023 Oct/Nov · Paper 22 · Q5 (b)5 marksSolving cubic equations
  • 2023 May/June · Paper 22 · Q3 (a, b)8 marksFactor and remainder theorems · Solving cubic equations

2022

  • 2022 Oct/Nov · Paper 22 · Q55 marksSolving cubic equations
  • 2022 May/June · Paper 12 · Q6 (a, b)5 marksFactor and remainder theorems

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Solving cubic equations · Core · Paper 1 style · no calculator · 4 marks
  1. (a) Show that x = 2 is a root of the equation x³ − 3x² − 4x + 12 = 0. [1]
  2. (b) Hence solve x³ − 3x² − 4x + 12 = 0. [3]
Show answer
  1. (a) Shown
    8 − 12 − 8 + 12 = 0 B1
  2. (b) x = 2, −2, 3
    Divide by (x − 2): x² − x − 6 M1 A1; (x + 2)(x − 3) → x = −2, 3 A1
Q2Solving cubic equations · Core · Paper 1 style · no calculator · 4 marks
  1. (a) Do not use a calculator in this question.Solve 2x³ + 3x² − 11x − 6 = 0. [4]
Show answer
  1. (a) x = 2, −½, −3
    Trial: f(2) = 16 + 12 − 22 − 6 = 0, so (x − 2) is a factor M1 A1; quotient 2x² + 7x + 3 = (2x + 1)(x + 3) M1; x = 2, −½, −3 A1
Q3Solving cubic equations · Core · Paper 1 style · no calculator · 4 marks
  1. (a) Do not use a calculator in this question.Find the exact roots of 4x³ − 13x − 6 = 0. [4]
Show answer
  1. (a) x = 2, −½, −3/2
    f(2) = 32 − 26 − 6 = 0 M1 A1; 4x³ − 13x − 6 = (x − 2)(4x² + 8x + 3) M1; (2x + 1)(2x + 3) → x = −½, −3/2 A1
Q4Solving cubic equations · Core · Paper 1 style · no calculator · 4 marks
  1. (a) Do not use a calculator in this question.Solve x³ − 4x² − 2x + 8 = 0, giving the roots in exact form. [4]
Show answer
  1. (a) x = 4, √2, −√2
    x²(x − 4) − 2(x − 4) M1; (x − 4)(x² − 2) = 0 A1; x = 4 A1; x = ±√2 A1
Q5Solving cubic equations · Core · Paper 1 style · no calculator · 5 marks
  1. (a) Find the x-coordinates of the points where the curve y = 2x³ − 5x² − 4x + 3 crosses the x-axis. [4]
  2. (b) State the coordinates of the point where the curve crosses the y-axis. [1]
Show answer
  1. (a) x = 3, ½, −1
    f(3) = 54 − 45 − 12 + 3 = 0 M1 A1; (x − 3)(2x² + x − 1) M1; (2x − 1)(x + 1) → x = ½, −1 A1
  2. (b) (0, 3)
    B1
Q6Solving cubic equations · Standard · Paper 1 style · no calculator · 6 marks
  1. (a) Factorise 3x³ − 10x² + 9x − 2 completely. [3]
  2. (b) Hence solve 3y⁶ − 10y⁴ + 9y² − 2 = 0, giving exact answers. [3]
Show answer
  1. (a) (x − 1)(3x − 1)(x − 2)
    f(1) = 0 so (x − 1) is a factor B1; quotient 3x² − 7x + 2 M1; (3x − 1)(x − 2) A1
  2. (b) y = ±1, ±1/√3 (= ±√3/3), ±√2
    Put x = y² M1; y² = 1, ⅓, 2 A1; y = ±1, ±√3/3, ±√2 A1
Q7Solving cubic equations · Standard · Paper 1 style · no calculator · 4 marks
  1. (a) Do not use a calculator in this question.Given that x = 3 is a root of x³ − 5x² + 5x + 3 = 0, find the other two roots in the form a ± √b. [4]
Show answer
  1. (a) x = 1 + √2, 1 − √2
    (x − 3)(x² − 2x − 1) M1 A1; x = (2 ± √8)/2 M1; = 1 ± √2 A1
Q8Solving cubic equations · Standard · Paper 1 style · no calculator · 6 marks

The curve y = x³ − 2x² − 3x + 5 and the line y = 2x − 1 intersect at three points.

  1. (a) Show that the x-coordinates of the points of intersection satisfy x³ − 2x² − 5x + 6 = 0. [1]
  2. (b) Find the coordinates of the three points of intersection. [5]
Show answer
  1. (a) Shown
    x³ − 2x² − 3x + 5 = 2x − 1 → x³ − 2x² − 5x + 6 = 0 B1
  2. (b) (1, 1), (−2, −5), (3, 5)
    f(1) = 0 B1; (x − 1)(x² − x − 6) M1 A1; x = −2, 3 A1; y-values from y = 2x − 1 A1
Q9Solving cubic equations · Standard · Paper 2 style · calculator · 6 marks

A storage box in a Karachi warehouse is a cuboid with edges x cm, (x + 2) cm and (x − 1) cm. Its volume is 72 cm³.

  1. (a) Show that x³ + x² − 2x − 72 = 0. [2]
  2. (b) Show that x = 4 is the only real solution of this equation, and state the dimensions of the box. [4]
Show answer
  1. (a) Shown
    x(x + 2)(x − 1) = x³ + x² − 2x M1; = 72 A1
  2. (b) x = 4; box is 4 cm × 6 cm × 3 cm
    f(4) = 64 + 16 − 8 − 72 = 0 B1; (x − 4)(x² + 5x + 18) M1 A1; discriminant 25 − 72 = −47 < 0 so no other real roots; 4, 6, 3 cm A1
Q10Solving cubic equations · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) Factorise x³ − 3x² − 9x − 5 completely. [3]
  2. (b) Hence state the roots of x³ − 3x² − 9x − 5 = 0, and describe what happens to the curve y = x³ − 3x² − 9x − 5 at x = −1. [2]
Show answer
  1. (a) (x + 1)²(x − 5)
    f(−1) = −1 − 3 + 9 − 5 = 0 B1; quotient x² − 4x − 5 M1; (x + 1)(x − 5) so (x + 1)²(x − 5) A1
  2. (b) x = −1 (repeated), x = 5; the curve touches the x-axis at (−1, 0)
    B1 roots; B1 touches (repeated root)
Q11Solving cubic equations · Standard · Paper 1 style · no calculator · 6 marks
  1. (a) Find the coordinates of the points where the curve y = x³ + 2x² meets the line y = 5x + 6. [6]
Show answer
  1. (a) (−1, 1), (2, 16), (−3, −9)
    x³ + 2x² − 5x − 6 = 0 M1; f(−1) = 0 B1; (x + 1)(x² + x − 6) M1; (x + 1)(x − 2)(x + 3) A1; x = −1, 2, −3 A1; y = 1, 16, −9 A1
Q12Solving cubic equations · Standard · Paper 2 style · calculator · 6 marks

A small business in Faisalabad models its monthly profit, P thousand rupees, t months after opening by P = −t³ + 9t² − 23t + 15, for 0 ≤ t ≤ 6.

  1. (a) Show that P = 0 when t = 1. [1]
  2. (b) Find the other values of t for which P = 0. [3]
  3. (c) Find the profit when t = 4, and state whether the business is making a profit or a loss at that time. [2]
Show answer
  1. (a) Shown
    −1 + 9 − 23 + 15 = 0 B1
  2. (b) t = 3, t = 5
    −(t − 1)(t² − 8t + 15) M1 A1; (t − 3)(t − 5) → t = 3, 5 A1
  3. (c) P = 3, a profit of 3000 rupees
    −64 + 144 − 92 + 15 = 3 M1 A1
Q13Solving cubic equations · Standard · Paper 1 style · no calculator · 4 marks
  1. (a) Do not use a calculator in this question.Solve 6x³ + 7x² − x − 2 = 0. [4]
Show answer
  1. (a) x = −1, −⅔, ½
    f(−1) = −6 + 7 + 1 − 2 = 0 M1 A1; quotient 6x² + x − 2 = (3x + 2)(2x − 1) M1; x = −1, −⅔, ½ A1
Q14Solving cubic equations · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) Factorise 2x³ − x² − 8x + 4 completely. [3]
  2. (b) Hence solve 2e³ʸ − e²ʸ − 8eʸ + 4 = 0, giving your answers in exact form. [3]
Show answer
  1. (a) (x − 2)(x + 2)(2x − 1)
    x²(2x − 1) − 4(2x − 1) M1; (2x − 1)(x² − 4) A1; (2x − 1)(x − 2)(x + 2) A1
  2. (b) y = ln 2, y = −ln 2 (= ln ½)
    eʸ = 2, ½ or −2 M1; eʸ = −2 has no solution B1; y = ln 2, ln ½ A1
Q15Solving cubic equations · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) Show that (x − 1) is a factor of p(x) = x³ − 6x² + 11x − 6 and factorise p(x) completely. [3]
  2. (b) Hence solve 8ᵘ − 6(4ᵘ) + 11(2ᵘ) − 6 = 0, giving your answers in exact form. [3]
Show answer
  1. (a) (x − 1)(x − 2)(x − 3)
    p(1) = 1 − 6 + 11 − 6 = 0 B1; quotient x² − 5x + 6 M1; (x − 2)(x − 3) A1
  2. (b) u = 0, u = 1, u = log₂ 3
    Put x = 2ᵘ, so 8ᵘ = x³ and 4ᵘ = x² M1; 2ᵘ = 1, 2 or 3 A1; u = 0, 1, log₂ 3 A1
Q16Solving cubic equations · Challenge · Paper 2 style · calculator · 7 marks

The point P(1, −2) lies on the curve y = x³ − 4x + 1.

  1. (a) Find the equation of the tangent to the curve at P. [3]
  2. (b) The tangent meets the curve again at Q. Find the coordinates of Q. [4]
Show answer
  1. (a) y = −x − 1
    dy/dx = 3x² − 4 M1; at x = 1 gradient −1 A1; y + 2 = −(x − 1) → y = −x − 1 A1
  2. (b) Q(−2, 1)
    x³ − 4x + 1 = −x − 1 → x³ − 3x + 2 = 0 M1; (x − 1)²(x + 2) = 0 M1 A1 (repeated root at P); Q(−2, 1) A1
Q17Solving cubic equations · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) Do not use a calculator in this question.By first solving 2t³ − 3t² − 3t + 2 = 0, solve the equation 2(ln x)³ − 3(ln x)² − 3 ln x + 2 = 0, giving your answers in exact form. [6]
Show answer
  1. (a) x = e², x = √e (= e^½), x = 1/e
    f(2) = 16 − 12 − 6 + 2 = 0 M1 A1; (t − 2)(2t² + t − 1) = (t − 2)(2t − 1)(t + 1) M1 A1; ln x = 2, ½, −1 M1; x = e², e^½, e⁻¹ A1
Q18Factor and remainder theorems · Core · Paper 1 style · no calculator · 4 marks
  1. (a) Find the remainder when 2x³ − 3x² + 5x − 7 is divided by (x − 2). [2]
  2. (b) Find the remainder when x³ + 2x² − 5 is divided by (x + 3). [2]
Show answer
  1. (a) 7
    f(2) = 16 − 12 + 10 − 7 M1; = 7 A1
  2. (b) −14
    g(−3) = −27 + 18 − 5 M1; = −14 A1
Q19Factor and remainder theorems · Core · Paper 1 style · no calculator · 2 marks
  1. (a) Find the remainder when 4x³ + 2x² − 3x + 5 is divided by (2x + 1). [2]
Show answer
  1. (a) 13/2 (6.5)
    f(−½) = −½ + ½ + 3/2 + 5 M1; = 13/2 A1
Q20Factor and remainder theorems · Core · Paper 1 style · no calculator · 5 marks

f(x) = 4x³ − 8x² + kx + 3, where k is a constant. (2x + 1) is a factor of f(x).

  1. (a) Show that k = 1. [2]
  2. (b) Factorise f(x) completely. [3]
Show answer
  1. (a) Shown
    f(−½) = −½ − 2 − k/2 + 3 = 0 M1; k = 1 A1
  2. (b) (2x + 1)(2x − 3)(x − 1)
    Quotient 2x² − 5x + 3 M1 A1; (2x − 3)(x − 1) A1
Q21Factor and remainder theorems · Standard · Paper 1 style · no calculator · 7 marks

p(x) = x³ + ax² + bx − 12, where a and b are constants. (x − 2) is a factor of p(x), and when p(x) is divided by (x + 1) the remainder is −18.

  1. (a) Find the values of a and b. [4]
  2. (b) Show that the equation p(x) = 0 has only one real root. [3]
Show answer
  1. (a) a = −1, b = 4
    p(2) = 0: 8 + 4a + 2b − 12 = 0 → 2a + b = 2 M1; p(−1) = −18: −1 + a − b − 12 = −18 → a − b = −5 M1; solve A1 A1
  2. (b) Shown: p(x) = (x − 2)(x² + x + 6), discriminant −23 < 0
    Quotient x² + x + 6 M1 A1; 1 − 24 = −23 < 0 so only x = 2 A1
Q22Factor and remainder theorems · Standard · Paper 1 style · no calculator · 7 marks

f(x) = 2x³ + px² + qx + 6. (x + 2) is a factor of f(x), and the remainder when f(x) is divided by (x − 1) is −6.

  1. (a) Find the values of p and q. [4]
  2. (b) Factorise f(x) completely. [3]
Show answer
  1. (a) p = −3, q = −11
    f(−2) = 0: −16 + 4p − 2q + 6 = 0 → 2p − q = 5 M1; f(1) = −6: 2 + p + q + 6 = −6 → p + q = −14 M1; p = −3 A1, q = −11 A1
  2. (b) (x + 2)(2x − 1)(x − 3)
    Divide by (x + 2): 2x² − 7x + 3 M1 A1; (2x − 1)(x − 3) A1
Q23Factor and remainder theorems · Standard · Paper 1 style · no calculator · 6 marks
  1. (a) Show that (2x − 1) is a factor of 2x³ + 5x² − 13x + 5. [2]
  2. (b) Hence solve 2x³ + 5x² − 13x + 5 = 0, giving the irrational roots in exact form. [4]
Show answer
  1. (a) Shown
    f(½) = ¼ + 5/4 − 13/2 + 5 M1; = 0 A1
  2. (b) x = ½, x = (−3 ± √29)/2
    Quotient x² + 3x − 5 M1 A1; x = (−3 ± √(9 + 20))/2 M1 A1
Q24Factor and remainder theorems · Standard · Paper 1 style · no calculator · 6 marks

f(x) = ax³ + bx² + 3x − 2, where a and b are constants. When f(x) is divided by (x − 1) the remainder is 5, and when f(x) is divided by (x + 2) the remainder is −16.

  1. (a) Find the values of a and b. [4]
  2. (b) Find the remainder when f(x) is divided by (2x − 1). [2]
Show answer
  1. (a) a = 2, b = 2
    a + b + 1 = 5 → a + b = 4 M1; −8a + 4b − 8 = −16 → 2a − b = 2 M1; a = 2, b = 2 A1 A1
  2. (b) ¼
    f(½) = 2/8 + 2/4 + 3/2 − 2 M1; = ¼ A1
Q25Factor and remainder theorems · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) Both (x − 1) and (x − 3) are factors of x³ + ax² + 11x + b. Find the values of a and b, and the third factor. [5]
Show answer
  1. (a) a = −6, b = −6; third factor (x − 2)
    1 + a + 11 + b = 0 → a + b = −12 M1; 27 + 9a + 33 + b = 0 → 9a + b = −60 M1; a = −6, b = −6 A1 A1; third factor: constant term −6 = (−1)(−3)(−c) gives (x − 2) A1
Q26Factor and remainder theorems · Standard · Paper 1 style · no calculator · 5 marks

f(x) = x³ + ax² − 4x + 2. The remainder when f(x) is divided by (x + 1) is three times the remainder when f(x) is divided by (x − 1).

  1. (a) Find the value of a. [3]
  2. (b) Find the remainder when f(x) is divided by (x + 2). [2]
Show answer
  1. (a) a = 4
    f(−1) = a + 5, f(1) = a − 1 B1; a + 5 = 3(a − 1) M1; a = 4 A1
  2. (b) 18
    f(−2) = −8 + 16 + 8 + 2 M1; = 18 A1
Q27Factor and remainder theorems · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) (x + 1) and (2x − 1) are both factors of 2x³ + ax² + bx − 3. Find the values of a and b and hence write the cubic as a product of three linear factors. [5]
Show answer
  1. (a) a = 7, b = 2; (x + 1)(2x − 1)(x + 3)
    f(−1) = 0: −2 + a − b − 3 = 0 → a − b = 5 M1; f(½) = 0: ¼ + a/4 + b/2 − 3 = 0 → a + 2b = 11 M1; a = 7, b = 2 A1; comparing constants gives third factor (x + 3) M1 A1
Q28Factor and remainder theorems · Challenge · Paper 1 style · no calculator · 7 marks

p(x) = 3x³ + ax² + bx + c. The remainder when p(x) is divided by x is −4. (x − 1) is a factor of p(x). The remainder when p(x) is divided by (x + 1) is −6.

  1. (a) Find the values of a, b and c. [4]
  2. (b) Show that x = 1 is the only real solution of p(x) = 0. [3]
Show answer
  1. (a) a = 1, b = 0, c = −4
    Divide by x means p(0) = c = −4 B1; p(1) = 0: a + b = 1 M1; p(−1) = −6: a − b = 1 M1; a = 1, b = 0 A1
  2. (b) Shown
    p(x) = (x − 1)(3x² + 4x + 4) M1 A1; discriminant 16 − 48 < 0 A1
Q29Factor and remainder theorems · Challenge · Paper 1 style · no calculator · 9 marks

f(x) = x³ + kx² − 2kx − 8, where k is a constant. When f(x) is divided by (x − k) the remainder is 28.

  1. (a) Show that k³ − k² − 18 = 0. [2]
  2. (b) Show that k = 3 is the only real value of k. [3]
  3. (c) Using k = 3, show that (x + 1) is a factor of f(x) and hence solve f(x) = 0. [4]
Show answer
  1. (a) Shown
    f(k) = k³ + k³ − 2k² − 8 = 28 M1; 2k³ − 2k² − 36 = 0 → k³ − k² − 18 = 0 A1
  2. (b) Shown
    k = 3: 27 − 9 − 18 = 0 B1; (k − 3)(k² + 2k + 6) M1; discriminant 4 − 24 < 0 A1
  3. (c) x = −1, −4, 2
    f(x) = x³ + 3x² − 6x − 8; f(−1) = −1 + 3 + 6 − 8 = 0 B1; quotient x² + 2x − 8 M1 A1; (x + 4)(x − 2) → x = −1, −4, 2 A1
Q30Factor and remainder theorems · Challenge · Paper 1 style · no calculator · 8 marks

g(x) = x³ + ax² + bx − 6. g(x) leaves the same remainder when divided by (x − 1) as when divided by (x + 2). Also, (x + 3) is a factor of g(x).

  1. (a) Show that a − b = 3. [2]
  2. (b) Find the values of a and b. [3]
  3. (c) Factorise g(x) completely and state the common remainder. [3]
Show answer
  1. (a) Shown
    g(1) = a + b − 5, g(−2) = 4a − 2b − 14 M1; equate: 3a − 3b = 9 → a − b = 3 A1
  2. (b) a = 4, b = 1
    g(−3) = 0: −27 + 9a − 3b − 6 = 0 → 3a − b = 11 M1; solve with a − b = 3 M1; a = 4, b = 1 A1
  3. (c) (x + 3)(x + 2)(x − 1); remainder 0
    Quotient x² + x − 2 M1 A1; (x + 2)(x − 1), so g(1) = g(−2) = 0 A1

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Solving cubic equations Banker10 questions · 8/9 sessions
Factor and remainder theorems Banker8 questions · 8/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Solving cubic equations Banker398/9
Factor and remainder theorems Banker318/9

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