Revision mat · Factor Theorem

14 marks

Question mix weighted by 33 real exam questions from 18 GCSE papers (2017–24), so the most common question types get the most space. Show your working.

Factor theorem $f(a)=0\iff(x-a)\text{ is a factor}$Remainder theorem $f(a)=\text{remainder on dividing by }(x-a)$Linear factor $(ax-b)$ $f\!\left(\dfrac ba\right)=0$Find an unknown $\text{substitute and solve}$Factorise a cubic $f(x)=(x-a)(\text{quadratic})$

Factorise & solve cubics≈55% of real exam Qs

1Solve $x^3 - 7x^2 + 14x - 8 = 0$, given that $(x - 1)$ is a factor.

Answer: (3 marks)

2Solve $x^3 - 7x - 6 = 0$, given that $(x + 2)$ is a factor.

Answer: (3 marks)

Factor & remainder theorem≈24% of real exam Qs

3Find the remainder when $f(x) = x^3 - 4x^2 + 2x - 1$ is divided by $(x - 2)$.

Answer: (2 marks)

Find an unknown coefficient≈9% of real exam Qs

4$(x + 3)$ is a factor of $f(x) = x^3 + 2x^2 + 2x + k$, where $k$ is a constant.
Find the value of $k$.

Answer: (3 marks)

★ Exam capstoneeverything else ≈12% of real exam Qs

5The graph shows $y = f(x)$, a cubic that crosses the $x$-axis at the three marked points.
The coefficient of $x^3$ in $f(x)$ is $1$.
Write $f(x)$ as a product of three linear factors.

Answer: (3 marks)

Mark scheme · Factor Theorem

Total: 14 marks

Award the marks shown for each correct step, then add up the total out of 14. A method mark counts even if the final answer is wrong.

1Solve $x^3 - 7x^2 + 14x - 8 = 0$, given that $(x - 1)$ is a factor.[3 marks]

Method

Divide by the factor: $x^3 - 7x^2 + 14x - 8 = (x - 1)(x^2 - 6x + 8) = 0$

Factorise the quadratic: $(x - 1)(x - 2)(x - 4) = 0$

So $x = 1,\ 2,\ 4$

Answer: $x = 1,\ 2,\ 4$

Marks

✔1 markdivide by the given factor

✔1 markfactorise the quadratic

✔1 markx = 1, 2, 4

2Solve $x^3 - 7x - 6 = 0$, given that $(x + 2)$ is a factor.[3 marks]

Method

Divide by the factor: $x^3 - 7x - 6 = (x + 2)(x^2 - 2x - 3) = 0$

Factorise the quadratic: $(x + 2)(x + 1)(x - 3) = 0$

So $x = -2,\ -1,\ 3$

Answer: $x = -2,\ -1,\ 3$

Marks

✔1 markdivide by the given factor

✔1 markfactorise the quadratic

✔1 markx = -2, -1, 3

3Find the remainder when $f(x) = x^3 - 4x^2 + 2x - 1$ is divided by $(x - 2)$.[2 marks]

Method

By the remainder theorem, the remainder is $f(2)$.

$f(2) = 8 - 16 + 4 - 1 = -5$

Answer: $-5$

Marks

✔1 markremainder = f(2)

✔1 mark= -5

4$(x + 3)$ is a factor of $f(x) = x^3 + 2x^2 + 2x + k$, where $k$ is a constant.
Find the value of $k$.[3 marks]

Method

By the factor theorem, $f(-3) = 0$.

$f(-3) = -27 + 18 - 6 + k = -15 + k$

$-15 + k = 0$, so $k = 15$

Answer: $15$

Marks

✔1 markuse f(-3) = 0

✔1 markform -15 + k = 0

✔1 markk = 15

5The graph shows $y = f(x)$, a cubic that crosses the $x$-axis at the three marked points.
The coefficient of $x^3$ in $f(x)$ is $1$.
Write $f(x)$ as a product of three linear factors.[3 marks]

Method

The curve crosses the $x$-axis at $x = -3$, $x = -2$ and $x = 3$.

Each crossing $x = r$ gives a factor $(x - r)$, so $f(x) = (x + 3)(x + 2)(x - 3)$.

Answer: $(x + 3)(x + 2)(x - 3)$

Marks

✔1 markread the roots: -3, -2, 3

✔1 markroot r → factor (x − r)

✔1 mark(x + 3)(x + 2)(x - 3)