What is the difference between the factor theorem and the remainder theorem?

The remainder theorem says p(a) is the remainder when p(x) is divided by (x - a). The factor theorem is the special case where that remainder is 0, which makes (x - a) a factor.

What value do I substitute when dividing by (2x - 1)?

Solve 2x - 1 = 0, so substitute x = 1/2. Substituting 1 or 2 is the most common error on this topic.

How do I find the quadratic factor after finding one linear factor?

Divide the cubic by the linear factor, by long division, synthetic division or by comparing coefficients. Then factorise or solve the quadratic that is left.

4037 Factor and Remainder Theorem Questions, Solved

The factor and remainder theorems appear on nearly every 4037 Paper 1 or Paper 2, often as a 6 to 8 mark question in two parts: find the unknown coefficients, then factorise or solve. The three questions below cover every version you are likely to meet.

Question 1: two unknowns, then solve [7 marks]

$p (x) = 2 x^{3} + a x^{2} - 13 x + b$ . $(x - 3)$ is a factor of $p (x)$ , and the remainder when $p (x)$ is divided by $(x + 1)$ is $12$ .

(a) Find the values of $a$ and $b$ .

(b) Solve $p (x) = 0$ , giving the non-integer roots in exact form.

Show the worked solution

(a) Each fact about $p$ gives one equation.

$(x - 3)$ is a factor, so $p (3) = 0$ :

54 + 9 a - 39 + b = 0 ⟹ 9 a + b = - 15

The remainder on dividing by $(x + 1)$ is $p (- 1)$ :

- 2 + a + 13 + b = 12 ⟹ a + b = 1

Subtract the second equation from the first: $8 a = - 16$ , so $a = - 2$ and $b = 3$ .

(b) $p (x) = 2 x^{3} - 2 x^{2} - 13 x + 3$ , and $(x - 3)$ is a factor. Dividing gives

p (x) = (x - 3) (2 x^{2} + 4 x - 1)

The quadratic does not factorise, so use the formula:

x = \frac{- 4 \pm 16 + 8}{4} = \frac{- 4 \pm 2 6}{4} = \frac{- 2 \pm 6}{2}

So $x = 3$ or $x = \frac{- 2 \pm 6}{2}$ .

Check: $p (- 1) = - 2 - 2 + 13 + 3 = 12$ .

Question 2: show a factor, then factorise completely [4 marks]

Show that $(x - 2)$ is a factor of $x^{3} - 3 x^{2} - 4 x + 12$ , and hence factorise it completely.

Show the worked solution

Substitute $x = 2$ :

8 - 12 - 8 + 12 = 0

The remainder is 0, so $(x - 2)$ is a factor. Say so in words: that sentence carries the mark.

Group the terms in pairs:

x^{2} (x - 3) - 4 (x - 3) = (x - 3) (x^{2} - 4) = (x - 3) (x - 2) (x + 2)

"Completely" means the difference of two squares $x^{2} - 4$ must be split too.

Question 3: remainder on dividing by (2x - 1) [2 marks]

Find the remainder when $4 x^{3} - 3 x + 5$ is divided by $(2 x - 1)$ .

Show the worked solution

$2 x - 1 = 0$ when $x = \frac{1}{2}$ , so the remainder is $p (\frac{1}{2})$ :

4 (\frac{1}{8}) - \frac{3}{2} + 5 = \frac{1}{2} - \frac{3}{2} + 5 = 4

The remainder is $4$ .

Try it: factorise any polynomial

Type a cubic, try to spot a root first, then compare with the working.

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O Level Additional Maths 4037: factor and remainder theorem questions

Question 1: two unknowns, then solve [7 marks]

Question 2: show a factor, then factorise completely [4 marks]

Question 3: remainder on dividing by (2x - 1) [2 marks]

Try it: factorise any polynomial

Questions students ask