The Factor and Remainder Theorems

Use the remainder and factor theorems to find remainders without dividing, show a bracket is a factor, and find unknown coefficients from given factors and remainders.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 26 real past-paper questions below

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Remainder without dividing

Long division works, but it is slow. Often a question only wants the remainder. There is a shortcut that finds it in one line, with no division at all.

Start from the division fact you already know, and put $x = a$ . Each line follows from the one before:

$P (x) = (x - a) Q (x) + R$ — divisor × quotient + remainder.

$P (a) = (a - a) Q (a) + R$ — put x = a everywhere.

$P (a) = 0 \times Q (a) + R$ — a − a is 0, so the quotient part vanishes.

$P (a) = R$ — only the remainder is left.

Formula

R = P (a)

$R$ — the remainder when P(x) is divided by $x - a$ .
$a$ — the number that makes the bracket $x - a$ equal to zero.

This is the remainder theorem. Learn it: no formula sheet prints it, on 0606, 4037 or 4PM1.

Worked example

1
Find the remainder when $x^{3} - 4 x^{2} + x + 7$ is divided by $x - 3$ .
2
$x - 3 = 0 \Rightarrow x = 3$ — Find the number that makes the bracket zero.
3
$P (3) = 27 - 36 + 3 + 7$ — Put 3 in place of every x: 3³ = 27 and 4 × 3² = 36.
4
$= 1$ — The remainder is 1. Long division gives the same answer, much more slowly.

Key idea

The remainder is just the value of the polynomial at the number that makes the bracket zero.

Brackets with a plus

For a bracket like $x + 2$ , ask: which number makes it zero? The answer is $- 2$ , so you work out $P (- 2)$ .

Worked example

1
Find the remainder when $x^{3} - 4 x^{2} + x + 7$ is divided by $x + 2$ .
2
$x + 2 = 0 \Rightarrow x = - 2$ — Set the bracket equal to zero.
3
$P (- 2) = (- 2)^{3} - 4 (- 2)^{2} + (- 2) + 7$ — Put −2 in brackets every time it replaces x.
4
$= - 8 - 16 - 2 + 7$ — (−2)³ = −8 and (−2)² = 4, so 4 × 4 = 16.
5
$= - 19$ — The remainder is −19.

Common mistake

Working out $P (2)$ for the bracket $x + 2$ . The plus sign means you use $- 2$ .

Common mistake

Writing $- 2^{2} = - 4$ . Without brackets the square acts on 2 only. Always write $(- 2)^{2} = 4$ .

Key idea

Set the bracket equal to zero and solve. That value is what goes into P.

Brackets like ax + b

Both boards also ask about brackets like $2 x - 1$ or $3 x + 2$ . The idea is the same: find the x that makes the bracket zero. This time it is a fraction.

Formula

R = P (- \frac{b}{a})

$R$ — the remainder when P(x) is divided by $a x + b$ .
$a$ — the number in front of x in the bracket.
$b$ — the number term in the bracket. Solving $a x + b = 0$ gives $x = - \frac{b}{a}$ .

Worked example

1
Find the remainder when $6 x^{3} + 7 x^{2} - 9 x + 5$ is divided by $2 x - 1$ .
2
$2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ — Solve to find the value to use.
3
$P (\frac{1}{2}) = 6 (\frac{1}{8}) + 7 (\frac{1}{4}) - 9 (\frac{1}{2}) + 5$ — One half cubed is one eighth; one half squared is one quarter.
4
$= \frac{3}{4} + \frac{7}{4} - \frac{9}{2} + 5$ — Multiply out each term.
5
$= \frac{10}{4} - \frac{18}{4} + \frac{20}{4} = 3$ — Write everything in quarters, then add: 12 quarters is 3.

Common mistake

Using $P (1)$ or $P (2)$ for the bracket $2 x - 1$ . You need the value that makes the whole bracket zero: one half.

Rule

0606 Paper 1 has no calculator. Keep fractions as fractions and use a common denominator. On 4PM1 a calculator is allowed, but show the substituted line.

Key idea

Any bracket, same question: what x makes it zero? Fractions are fine.

The factor theorem

A factor divides exactly, so its remainder is 0. Put $R = 0$ into the remainder theorem and you get the factor theorem.

Formula

P (a) = 0 ⟺ (x - a) is a factor of P (x)

$a$ — the number that makes the bracket zero.
$⟺$ — works both ways: if one side is true, so is the other.

Formula

P (- \frac{b}{a}) = 0 ⟺ (a x + b) is a factor of P (x)

$a$ — the number in front of x in the bracket.
$b$ — the number term in the bracket.

Worked example

1
Show that $x - 2$ is a factor of $x^{3} + x^{2} - 4 x - 4$ .
2
$P (2) = 8 + 4 - 8 - 4$ — Put x = 2 and write every term out.
3
$= 0$ — The value is zero.
4
$So (x - 2) is a factor.$ — Finish with a sentence that names the theorem's conclusion.

Worked example

1
Given that $2 x + 1$ is a factor of $2 x^{3} + k x^{2} - 8 x - 3$ , find k.
2
$2 x + 1 = 0 \Rightarrow x = - \frac{1}{2}$ — The value that makes the factor zero.
3
$2 (- \frac{1}{8}) + k (\frac{1}{4}) - 8 (- \frac{1}{2}) - 3 = 0$ — A factor, so P(−½) = 0.
4
$- \frac{1}{4} + \frac{k}{4} + 4 - 3 = 0$ — Tidy each term.
5
$\frac{k}{4} = - \frac{3}{4}, k = - 3$ — Collect the numbers, then multiply both sides by 4.

Common mistake

In a show-that question, writing only $P (2) = 0$ . The mark is for the substituted values. Show them, then state the conclusion.

Key idea

Factor means remainder 0. So a factor gives you an equation equal to zero.

Two unknowns

The most common exam question has two unknown letters and two facts. Each fact gives one equation. Then you solve them together, like any pair of simultaneous equations.

Steps

1
1. Turn each fact into $P (value) =$ something: 0 for a factor, the remainder otherwise.
2. Simplify each one into a short linear equation in the two letters.
3. Solve the pair simultaneously.
4. Check both facts with your values.

Worked example

1
$P (x) = 2 x^{3} + a x^{2} + b x - 6$ . $(x - 2)$ is a factor, and dividing by $(x + 1)$ leaves remainder $- 12$ . Find a and b.
2
$P (2) = 16 + 4 a + 2 b - 6 = 0$ — Factor fact: put x = 2 and set it equal to 0.
3
$2 a + b = - 5$ — Tidy: 4a + 2b = −10, then halve every term.
4
$P (- 1) = - 2 + a - b - 6 = - 12$ — Remainder fact: put x = −1 and set it equal to −12.
5
$a - b = - 4$ — Tidy: a − b − 8 = −12.
6
$3 a = - 9, a = - 3$ — Add the two equations, so b cancels.
7
$b = 1$ — Put a = −3 into 2a + b = −5: −6 + b = −5.
8
$P (2) = 16 - 12 + 2 - 6 = 0$ — Check the factor fact with a = −3 and b = 1. It works.

Mark schemes give one mark for each correct equation, a method mark for solving them, and an accuracy mark for the values. So write the $P (2)$ and $P (- 1)$ lines in full.

Common mistake

Setting the remainder fact equal to 0. Only a factor gives 0. A remainder of $- 12$ means $P (- 1) = - 12$ .

Key idea

One fact, one equation. Two unknowns need two facts.

Equal remainders

Some questions link two remainders, for example: the remainder is the same for two different brackets. Write each remainder using the theorem, then set them equal.

Worked example

1
$x^{3} + k x^{2} - 2 x + 1$ leaves the same remainder when divided by $x - 1$ and by $x + 2$ . Find k.
2
$P (1) = 1 + k - 2 + 1 = k$ — First remainder: put x = 1.
3
$P (- 2) = - 8 + 4 k + 4 + 1 = 4 k - 3$ — Second remainder: put x = −2. Note k(−2)² = 4k and −2(−2) = +4.
4
$k = 4 k - 3$ — The remainders are equal.
5
$k = 1$ — Take k from both sides: 3k = 3.
6
$P (1) = 1, P (- 2) = - 8 + 4 + 4 + 1 = 1$ — Check: both remainders are 1.

Common mistake

Sign slips with negative values. Put every negative number in brackets before you square or cube it.

Key idea

Every fact about a factor or a remainder becomes one equation. Count the unknowns, then count the facts.

Which method? Read the question

Read the question first. The wording tells you which value to put in, and what to set it equal to.

You see	You do
Remainder when divided by $x - a$	Work out $P (a)$
Divided by $x + a$	Work out $P (- a)$
Divided by $a x + b$	Work out $P (- b / a)$ , a fraction
“Show that … is a factor”	Substitute, show the value is 0, then write the conclusion
“… is a factor”, find k	Set $P (value) = 0$ and solve for k
Two unknowns, two facts	Two equations, solve simultaneously
“the same remainder”	Write both remainders, set them equal

Common mistake

Expecting the theorem to give the quotient. It gives only the remainder. To find the quotient, you still divide.

Key idea

Bracket equal to zero, solve for x, put it into P. That one move answers every question in this lesson.

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