Remainder theorem | Year 12

01 / LEARN

The idea, step by step

Why substitution gives the remainder

When $P (x)$ is divided by $x - a$ , the division identity is $P (x) = (x - a) Q (x) + r$ . Put $x = a$ : the product vanishes and $P (a) = r$ . The remainder theorem avoids finding the entire quotient when only the remainder is needed.

For a non-monic linear divisor $u x - v$ , use its zero $a = v / u$ . Substituting that value still leaves the remainder, because the remainder is constant. Do not multiply the resulting value by the divisor leading coefficient.

Working backwards from remainder conditions

A statement such as “the remainder on division by $x - 1$ is $5$ ” means $P (1) = 5$ . Several conditions become a small system of equations in the unknown coefficients. After solving, substitute back into every original condition.

For a divisor $(x - a) (x - b)$ with distinct $a, b$ , the remainder has the form $m x + c$ . Its values at $a$ and $b$ match those of $P$ . Two substitutions therefore determine this linear remainder without a complete division.

A common trap

To divide by $2 x - 1$ , substitute $x = 2$ .

Instead: Use the zero of the divisor: $2 x - 1 = 0$ gives $x = \frac{1}{2}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

A non-monic divisor

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

The divisor is zero at $x = \frac{1}{2}$ .
$P (\frac{1}{2}) = 3 (\frac{1}{8}) - 1 + 7 = \frac{3}{8} + 6 = \frac{51}{8}$ .

\frac{51}{8}

Example 2

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 $a, b$ .

$P (1) = 5$ gives $1 + a + b = 5$ , so $a + b = 4$ .
$P (- 1) = - 1$ gives $- 1 - a + b = - 1$ , so $- a + b = 0$ .
Therefore $a = b = 2$ . Both original substitutions check.

a = 2, b = 2

Check your understanding

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

Show answer

$0$ .

Substitute $1$ : $1 + 2 - 3 = 0$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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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