A-level · Algebra & functions

Polynomials and the factor theoremExam-style questions

Solving a cubic by finding one root first. Which whole numbers are worth trying, and why; the remainder theorem, which finds a remainder with one substitution; the factor theorem, and testing a factor like (2x − 1); dividing out a factor by matching coefficients, with a 0 for every missing power; finishing the cubic from the quadratic that is left; and reading a sketch from the factors, where a squared factor touches the axis.

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35 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
(a) Show that $2x^3 - 3x^2 - 11x + 6$ can be written as $(x - 3)\left(2x^2 + 3x - 2\right)$ . (3 marks)
(b) Hence solve the equation $2x^3 - 3x^2 - 11x + 6 = 0$ . (2 marks)
Level 3 · 5 marksTry it
2
The polynomial $\mathrm{p}(x)$ is defined by

$\mathrm{p}(x) = 4x^{3} + ax^{2} - 11x + 6,$

where $a$ is a constant. It is given that $(2x - 1)$ is a factor of $\mathrm{p}(x)$ .
(a) Find the value of $a$ .

(2 marks)
(b) When $a$ has this value, find the remainder when $\mathrm{p}(x)$ is divided by $(x + 2)$ .

(2 marks)
Level 3 · 4 marksTry it
3
Solve the equation $x^{4} - 5x^{2} + 4 = 0$ . (4 marks)
Level 3 · 4 marksTry it
4
$\mathrm{f}(x) = 2x^3 + ax^2 - 5x + b$

where $a$ and $b$ are constants.

When $\mathrm{f}(x)$ is divided by $(x - 1)$ the remainder is $-6$ , and $(x + 2)$ is a factor of $\mathrm{f}(x)$ .
(a) Find the value of $a$ and the value of $b$ . (4 marks)
(b) Factorise $\mathrm{f}(x)$ completely. (2 marks)
Level 3 · 6 marksTry it
5
The polynomial $\text{p}(x)$ is defined by

$\text{p}(x) = x^3 + (b - 3)x^2 + (2 - 3b)x - 6$

where $b$ can take any real value.

Use the factor theorem to prove that $(x - 3)$ is always a factor of $\text{p}(x)$

(3 marks)
Level 3 · 3 marksTry it