Remainder theorem | Year 12

01 / LEARN

Understand the idea.

Why substitution gives the remainder

When dividing by $x - a$ , the remainder is a constant $r$ . Substituting $x = a$ into $P (x) = (x - a) Q (x) + r$ makes the product vanish and leaves $r = P (a)$ . This is the remainder theorem.

For a linear divisor $u x - v$ with $u \neq 0$ , substitute its zero $v / u$ . The remainder is still exactly that polynomial value. No additional multiplication or division by $u$ is needed.

Working backwards from remainder conditions

A supplied remainder becomes an equation in the unknown coefficients. For example, a remainder of five on division by $x - 1$ means $P (1) = 5$ . Translate each condition before solving the resulting system.

For a quadratic divisor with two distinct real zeros $a, b$ , the remainder has form $m x + c$ . Its values must match $P (a)$ and $P (b)$ , giving two equations. Check the solved coefficients in every original condition.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A non-monic divisor

Find the remainder when $P (x) = 3 x^{3} - 2 x + 7$ is divided by $2 x - 1$ .

Worked solution

1
Find the zero of the divisor.
$2 x - 1 = 0 \Rightarrow x = \frac{1}{2} .$
At this input, the divisor-times-quotient term is zero.
2
Use the remainder theorem.
$r = P (\frac{1}{2}) = 3 {(\frac{1}{2})}^{3} - 2 (\frac{1}{2}) + 7.$
A linear divisor leaves a constant remainder, equal to this value.
3
Evaluate exactly.
$\begin{aligned} r & = \frac{3}{8} - 1 + 7 \\ = \frac{3}{8} + 6 = \frac{51}{8} . \end{aligned}$
Combining fractions keeps the answer exact; the divisor coefficient does not rescale it.

Answer

$\frac{51}{8}$

EXAMPLE 02

Two supplied remainders

$P (x) = x^{3} + a x + b$ has remainder $5$ on division by $x - 1$ , and remainder $- 1$ on division by $x + 1$ . Find $a, b$ .

Worked solution

1
Translate the first remainder.
$P (1) = 5 : 1 + a + b = 5 \Rightarrow a + b = 4.$
The divisor $x - 1$ vanishes at one.
2
Translate the second remainder.
$P (- 1) = - 1 : - 1 - a + b = - 1 \Rightarrow - a + b = 0.$
The divisor $x + 1$ vanishes at negative one.
3
Solve the two equations.
$\begin{aligned} (a + b) + (- a + b) & = 4, \\ 2 b & = 4 \Rightarrow b = 2, \\ a & = 4 - b = 2. \end{aligned}$
Adding eliminates $a$ ; substituting then determines it.
4
Check both supplied conditions.
$\begin{aligned} P (1) & = 1 + 2 + 2 = 5, \\ P (- 1) & = - 1 - 2 + 2 = - 1. \end{aligned}$
Both original remainders are recovered for $a = 2, b = 2$ .

Answer

$a = 2, b = 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the remainder when $x^{5} + 2 x - 3$ is divided by $x - 1$ ?

Show answer & reasoning

$0$ .

The divisor $x - 1$ vanishes at one, so the remainder is $P (1) = 1^{5} + 2 (1) - 3 = 0$ . A zero remainder also means $x - 1$ is a factor.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the remainder when $x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 2$ .
2 marks
Find the remainder when $2 x^{3} + x^{2} - 5$ is divided by $x + 2$ .
2 marks
Find the remainder when $4 x^{3} - 2 x + 3$ is divided by $2 x + 1$ .
2 marks
$P (x) = x^{3} + k x + 2$ has remainder $8$ on division by $x - 2$ . Find $k$ .
2 marks
$P (1) = 4$ and $P (- 1) = 2$ . Find the remainder on dividing $P$ by $x^{2} - 1$ .
3 marks
$P (x)$ has remainder $3$ when divided by $x - 2$ . What remainder does $P (x) + 5 {(x - 2)}^{2} - 1$ have on division by $x - 2$ ?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the remainder when $x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 2$ .
2 marks
$P (2) = 16 - 12 + 4 + 1 = 9$ .
$9$
Find the remainder when $2 x^{3} + x^{2} - 5$ is divided by $x + 2$ .
2 marks
$P (- 2) = - 16 + 4 - 5 = - 17$ .
$- 17$
Find the remainder when $4 x^{3} - 2 x + 3$ is divided by $2 x + 1$ .
2 marks
At $x = - \frac{1}{2}$ , the value is $- \frac{1}{2} + 1 + 3 = \frac{7}{2}$ .
$\frac{7}{2}$
$P (x) = x^{3} + k x + 2$ has remainder $8$ on division by $x - 2$ . Find $k$ .
2 marks
$8 + 2 k + 2 = 8$ , so $2 k = - 2$ .
$k = - 1$
$P (1) = 4$ and $P (- 1) = 2$ . Find the remainder on dividing $P$ by $x^{2} - 1$ .
3 marks
Write $R = m x + c$ . Then $m + c = 4$ and $- m + c = 2$ , so $m = 1, c = 3$ .
$x + 3$
$P (x)$ has remainder $3$ when divided by $x - 2$ . What remainder does $P (x) + 5 {(x - 2)}^{2} - 1$ have on division by $x - 2$ ?
3 marks
At $x = 2$ , the added square is zero, so the new value is $3 - 1 = 2$ .
$2$

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Curriculum reference & lesson scope

SACE Board of South Australia

A recognisable real-polynomial chapter supporting Stage 2 Specialist polynomial factorisation and graph analysis; original lessons use current SACE scope and the school chapter structure requested by the user.

Official curriculum source ↗

In this lesson

Understand the idea.

Why substitution gives the remainder

Working backwards from remainder conditions

See the method in action.

A non-monic divisor

Find the zero of the divisor.

Use the remainder theorem.

Evaluate exactly.

Two supplied remainders

Translate the first remainder.

Translate the second remainder.

Solve the two equations.

Check both supplied conditions.

Try it yourself.

Check your understanding

Keep practising