Free Polynomials & roots questions for IB Mathematics: Analysis & Approaches (SL & HL), HL only, part of Functions. Each question has the answer and the marking points an examiner would look for.
Generate a full Polynomials & Roots worksheet →All Functions questionsThe roots of the equation x3 − 3x2 + kx + 3 = 0 form an arithmetic sequence.
Find the roots of the equation and the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| roots −1, 1, 3; k = −1 let the roots be α − d, α, α + d M1 sum of roots: 3α = 3 ⇒ α = 1 A1 product: α(α2 − d2) = −3 M1 d = ±2 A1 roots −1, 1, 3 A1 k = sum of pair products = −1 A1 | 6 |
Let P(z) = z4 + az3 + bz2 + cz + d, where a, b, c, d ∈ ℝ. Two of the roots of P(z) = 0 are 3 + 2i and −3.
Write down another root of P(z) = 0, giving a reason.
Find a quadratic factor of P(z) with real coefficients.
Given also that the sum of all four roots is 1, find the fourth root and hence the values of a, b, c and d.
Find the product of the four roots, and verify that it equals d.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 3 − 2i 3 − 2i A1 complex roots of polynomials with real coefficients occur in conjugate pairs R1 | 2 |
| (b) | z2 − 6z + 13 (z − (3 + 2i))(z − (3 − 2i)) M1 use of (z − p)2 + q2 (A1) z2 − 6z + 13 A1 | 3 |
| (c) | fourth root −2; a = −1, b = −11, c = 29, d = 78 2(3) − 3 + s = 1 M1 s = −2 A1 P(z) = (z2 − 6z + 13)(z2 + 5z + 6) M1 a = −1 A1 b = −11 A1 c = 29 A1 d = 78 A1 | 7 |
| (d) | 78 (13)(−3)(−2) M1 78 = d A1 | 2 |
Let p(x) = x3 + ax2 + bx + 15, where a, b ∈ ℝ. Given that (x − 3) is a factor of p(x), and that the remainder when p(x) is divided by (x + 2) is −35,
find the value of a and the value of b;
hence factorise p(x) fully.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | a = −7, b = 7 p(3) = 0: 9a + 3b = −42 M1A1 p(−2) = −35: 4a − 2b = −42 A1 attempt to solve simultaneously (M1) a = −7, b = 7 A1 | 5 |
| (b) | (x − 3)(x + 1)(x − 5) attempt to divide by the known factor (M1) (x − 3)(x + 1)(x − 5) A1 | 2 |
The roots of the equation x3 − 6x2 + kx − 6 = 0 form an arithmetic sequence.
Find the roots of the equation and the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| roots 1, 2, 3; k = 11 let the roots be α − d, α, α + d M1 sum of roots: 3α = 6 ⇒ α = 2 A1 product: α(α2 − d2) = 6 M1 d = ±1 A1 roots 1, 2, 3 A1 k = sum of pair products = 11 A1 | 6 |
Let P(z) = z4 + az3 + bz2 + cz + d, where a, b, c, d ∈ ℝ. Two of the roots of P(z) = 0 are −1 + i and −3.
Write down another root of P(z) = 0, giving a reason.
Find a quadratic factor of P(z) with real coefficients.
Given also that the sum of all four roots is −1, find the fourth root and hence the values of a, b, c and d.
Find the product of the four roots, and verify that it equals d.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −1 − i −1 − i A1 complex roots of polynomials with real coefficients occur in conjugate pairs R1 | 2 |
| (b) | z2 + 2z + 2 (z − (−1 + i))(z − (−1 − i)) M1 use of (z − p)2 + q2 (A1) z2 + 2z + 2 A1 | 3 |
| (c) | fourth root 4; a = 1, b = −12, c = −26, d = −24 2(−1) − 3 + s = −1 M1 s = 4 A1 P(z) = (z2 + 2z + 2)(z2 − z − 12) M1 a = 1 A1 b = −12 A1 c = −26 A1 d = −24 A1 | 7 |
| (d) | −24 (2)(−3)(4) M1 −24 = d A1 | 2 |
Let p(x) = x3 + ax2 + bx − 18, where a, b ∈ ℝ. Given that (x − 3) is a factor of p(x), and that the remainder when p(x) is divided by (x + 1) is −8,
find the value of a and the value of b;
hence factorise p(x) fully.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | a = 2, b = −9 p(3) = 0: 9a + 3b = −9 M1A1 p(−1) = −8: a − b = 11 A1 attempt to solve simultaneously (M1) a = 2, b = −9 A1 | 5 |
| (b) | (x − 3)(x + 3)(x + 2) attempt to divide by the known factor (M1) (x − 3)(x + 3)(x + 2) A1 | 2 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in