The Remainder Theorem

A2 · P3 · 12 min

Long division always works, but if all you want is the remainder after dividing by a linear expression, there is a far quicker route: substitute one number. That is the remainder theorem. It is the standard tool for finding unknown coefficients in a polynomial, and its special case, remainder zero, is the factor theorem. Typical P3 questions give two facts about a cubic (a factor, a remainder, or two remainders), ask you to find two unknown constants, and then go on to factorise or solve.

Why one substitution is enough

Divide $p(x)$ by $(x - a)$ . The remainder is a constant, call it $R$ , because a linear divisor leaves a remainder of degree $0$ . So for some quotient $q(x)$ ,

p(x) \equiv (x - a)\,q(x) + R.

This is an identity, true for every $x$ . Choose the one value that kills the first term, $x = a$ :

p(a) = (a - a)\,q(a) + R = 0 + R = R.

You do not need to know $q(x)$ at all. The remainder is simply $p(a)$ .

Key result

Remainder theorem. When a polynomial $p(x)$ is divided by $(x - a)$ , the remainder is $p(a)$ .

When $p(x)$ is divided by $(ax - b)$ , the remainder is $p\!\left(\dfrac{b}{a}\right)$ .

The (ax - b) version

Write $p(x) \equiv (ax - b)\,q(x) + R$ , with $R$ constant. Substitute $x = \dfrac{b}{a}$ , which makes $ax - b = 0$ :

p\!\left(\frac{b}{a}\right) = 0 \times q\!\left(\frac{b}{a}\right) + R = R.

The rule for which number to substitute: solve divisor $= 0$ . For $x + 3$ substitute $x = -3$ ; for $2x + 1$ substitute $x = -\tfrac{1}{2}$ ; for $3x - 2$ substitute $x = \tfrac{2}{3}$ .

Two remainders by substitution

Find the remainder when $p(x) = x^3 - 4x^2 + 2x + 7$ is divided by (a) $x - 3$ , (b) $2x + 1$ .

Solution

(a) $x - 3 = 0$ when $x = 3$ :

p(3) = 27 - 36 + 6 + 7 = 4.

The remainder is $4$ .

(b) $2x + 1 = 0$ when $x = -\tfrac{1}{2}$ :

p\!\left(-\tfrac{1}{2}\right) = -\tfrac{1}{8} - 4 \cdot \tfrac{1}{4} + 2\left(-\tfrac{1}{2}\right) + 7 = -\tfrac{1}{8} - 1 - 1 + 7 = \tfrac{39}{8}.

The remainder is $\tfrac{39}{8}$ .

Watch out

The remainder on dividing by $(2x + 1)$ is $p\!\left(-\tfrac{1}{2}\right)$ , not $p(-1)$ and not $2 \times$ anything. Substitute the root of the divisor, exactly as it is.

Finding unknown coefficients

Each piece of information ("the remainder is $8$ ", " $(x + 1)$ is a factor") becomes one equation in the unknowns. Two pieces of information, two unknowns, simultaneous equations.

Unknown coefficients from remainder and factor conditions

For each condition, substitute the root of the divisor into $p(x)$ .
Set the result equal to the given remainder (or to $0$ for a factor).
Simplify each equation to the form $\alpha a + \beta b = \gamma$ , clearing fractions.
Solve simultaneously.
Check by substituting the values back into one condition.

Two remainder conditions

The polynomial $p(x) = 2x^3 + ax^2 + bx - 4$ , where $a$ and $b$ are constants, leaves remainder $8$ when divided by $(x - 2)$ and remainder $-13$ when divided by $(x + 1)$ . Find $a$ and $b$ .

Solution

$p(2) = 8$ :

16 + 4a + 2b - 4 = 8 \quad\Rightarrow\quad 4a + 2b = -4 \quad\Rightarrow\quad 2a + b = -2.

$p(-1) = -13$ :

-2 + a - b - 4 = -13 \quad\Rightarrow\quad a - b = -7.

Add the two equations: $3a = -9$ , so $a = -3$ , and then $b = -2 - 2a = 4$ .

Check: $p(-1) = -2 - 3 - 4 - 4 = -13$ .

Equal remainders

The polynomial $x^3 + ax^2 - 5x + 2$ leaves the same remainder when divided by $(x - 2)$ and when divided by $(x + 1)$ . Find $a$ .

Solution

Let $p(x) = x^3 + ax^2 - 5x + 2$ .

$p(2) = 8 + 4a - 10 + 2 = 4a$ .

$p(-1) = -1 + a + 5 + 2 = a + 6$ .

The remainders are equal, so $4a = a + 6$ , giving $a = 2$ .

A non-monic divisor

When $4x^3 + kx^2 - 3x + 2$ is divided by $(2x + 1)$ the remainder is $5$ . Find $k$ .

Solution

Substitute $x = -\tfrac{1}{2}$ :

4\left(-\tfrac{1}{8}\right) + k \cdot \tfrac{1}{4} - 3\left(-\tfrac{1}{2}\right) + 2 = -\tfrac{1}{2} + \tfrac{k}{4} + \tfrac{3}{2} + 2 = 3 + \tfrac{k}{4}.

Set equal to $5$ : $\tfrac{k}{4} = 2$ , so $k = 8$ .

Combining a factor and a remainder

This is the most common exam shape: one condition says "is a factor" (so $p(\ldots) = 0$ ), the other gives a remainder. Afterwards you usually factorise, using the known factor.

Factor and remainder, then factorise

The polynomial $p(x) = ax^3 + bx^2 + 5x - 2$ , where $a$ and $b$ are constants, has $(2x - 1)$ as a factor. When $p(x)$ is divided by $(x + 1)$ the remainder is $-12$ .

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

(b) When $a$ and $b$ have these values, factorise $p(x)$ and show that the equation $p(x) = 0$ has only one real root.

Solution

(a) $(2x - 1)$ is a factor, so $p\!\left(\tfrac{1}{2}\right) = 0$ :

\frac{a}{8} + \frac{b}{4} + \frac{5}{2} - 2 = 0.

Multiply by $8$ : $a + 2b + 20 - 16 = 0$ , so $a + 2b = -4$ .

The remainder on division by $(x + 1)$ is $p(-1) = -12$ :

-a + b - 5 - 2 = -12 \quad\Rightarrow\quad -a + b = -5.

Add the equations: $3b = -9$ , so $b = -3$ ; then $a = -4 - 2b = 2$ .

(b) $p(x) = 2x^3 - 3x^2 + 5x - 2$ . Divide by $(2x - 1)$ , or compare coefficients in $p(x) \equiv (2x - 1)(x^2 + cx + 2)$ : the $x^2$ coefficient gives $2c - 1 = -3$ , so $c = -1$ . Check the $x$ coefficient: $4 - c = 5$ . So

p(x) = (2x - 1)(x^2 - x + 2).

The quadratic $x^2 - x + 2$ has discriminant $(-1)^2 - 4(1)(2) = -7 < 0$ , so it has no real roots. The only real root of $p(x) = 0$ is $x = \tfrac{1}{2}$ .

Exam tip

"Show that $p(x) = 0$ has only one real root" after factorising means: find the discriminant of the quadratic factor, show it is negative, and state the conclusion. A bare " $-7$ " without " $< 0$ , so no real roots" usually loses the final mark.

Remainders on dividing by a quadratic

If the divisor is a quadratic that factorises into two different linear factors, $(x - \alpha)(x - \beta)$ , the remainder has the form $Rx + S$ . Write

p(x) \equiv (x - \alpha)(x - \beta)\,q(x) + Rx + S

and substitute $x = \alpha$ and $x = \beta$ . Each kills the first term, leaving two equations for $R$ and $S$ . This is the remainder theorem extended.

Remainder on dividing by a quadratic

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

Solution

$x^2 - 1 = (x - 1)(x + 1)$ , so write

x^4 - 3x^2 + x + 5 \equiv (x^2 - 1)\,q(x) + Rx + S.

$x = 1$ : $1 - 3 + 1 + 5 = 4 = R + S$ .

$x = -1$ : $1 - 3 - 1 + 5 = 2 = -R + S$ .

Adding: $2S = 6$ , so $S = 3$ , and $R = 1$ . The remainder is $x + 3$ .

Remainder from information only

A polynomial $p(x)$ leaves remainder $5$ when divided by $(x - 1)$ and remainder $1$ when divided by $(x - 3)$ . Find the remainder when $p(x)$ is divided by $x^2 - 4x + 3$ .

Solution

You do not know $p(x)$ , but you do not need it. Since $x^2 - 4x + 3 = (x - 1)(x - 3)$ ,

p(x) \equiv (x - 1)(x - 3)\,q(x) + Rx + S.

The remainder theorem gives $p(1) = 5$ and $p(3) = 1$ . Substituting:

$x = 1$ : $R + S = 5$ .

$x = 3$ : $3R + S = 1$ .

Subtracting: $2R = -4$ , so $R = -2$ and $S = 7$ . The remainder is $7 - 2x$ .

Common mistakes

Watch out

Wrong sign in the substitution. Dividing by $(x + 2)$ means substituting $x = -2$ . The number to use is always the solution of "divisor $= 0$ ".

Watch out

Mixing up the two theorems. " $p(2) = 0$ , so $(x - 2)$ is a factor" (factor theorem) versus " $p(2) = 5$ , so the remainder is $5$ " (remainder theorem). If a question says "is a factor", the equation is $= 0$ .

Watch out

Arithmetic with negative powers of fractions. $\left(-\tfrac{1}{2}\right)^3 = -\tfrac{1}{8}$ and $\left(-\tfrac{1}{2}\right)^2 = +\tfrac{1}{4}$ . Write each term separately before adding, and multiply through by the common denominator early.

Exam tip

Show the substitution explicitly, for example " $p(-1) = -a + b - 5 - 2 = -12$ ". The method mark is for substituting the correct value and equating to the correct number.
Simplify each equation before solving: it is much easier to spot an error in $a + 2b = -4$ than in $\tfrac{a}{8} + \tfrac{b}{4} + \tfrac{1}{2} = 0$ .
After finding the constants, check one condition. A wrong value of $a$ ruins every later part.
If the next part says "hence factorise", divide by the factor you were given; do not search for a new one.

Summary

Dividing $p(x)$ by $(x - a)$ leaves remainder $p(a)$ ; dividing by $(ax - b)$ leaves remainder $p\!\left(\tfrac{b}{a}\right)$ .
Substitute the root of the divisor. Never divide when only the remainder is needed.
Each factor or remainder condition gives one linear equation in the unknown coefficients.
Factor $\Rightarrow$ $p(\ldots) = 0$ . Remainder $R$ $\Rightarrow$ $p(\ldots) = R$ .
For a quadratic divisor $(x - \alpha)(x - \beta)$ , the remainder is $Rx + S$ ; substitute $\alpha$ and $\beta$ .
To show only one real root, factorise and show the quadratic factor has negative discriminant.

Practice

Question

Find the remainder when $2x^3 + x^2 - 5x + 1$ is divided by $(x + 2)$ .
The polynomial $x^3 - 2x^2 + kx + 4$ leaves remainder $10$ when divided by $(x - 2)$ . Find $k$ .
Find the remainder when $9x^3 - 3x + 4$ is divided by $(3x - 2)$ .
The polynomial $2x^3 + ax^2 + bx - 3$ has $(x - 1)$ as a factor and leaves remainder $-15$ when divided by $(x + 2)$ . Find $a$ and $b$ .
The polynomial $x^3 + kx^2 + 2$ leaves the same remainder when divided by $(x - 1)$ and by $(x + 3)$ . Find $k$ .
Find the remainder when $x^3 + 2x^2 - x + 1$ is divided by $x^2 - x - 2$ .
The polynomial $p(x) = x^3 + ax + b$ leaves the same remainder when divided by $(x - 1)$ as when divided by $(x + 1)$ , and $(x - 2)$ is a factor of $p(x)$ . (a) Find $a$ and $b$ . (b) Show that $p(x) = 0$ has exactly one real root.
A polynomial $p(x)$ leaves remainder $5$ on division by $(x - 1)$ and remainder $1$ on division by $(x - 3)$ . Find the remainder when $p(x)$ is divided by $(x - 1)(x - 3)$ , and hence find the remainder when $p(x) + 2x$ is divided by $(x - 1)(x - 3)$ .

Answers

$p(-2) = 2(-8) + 4 + 10 + 1 = -1$ .
$p(2) = 8 - 8 + 2k + 4 = 2k + 4 = 10$ , so $k = 3$ .
$p\!\left(\tfrac{2}{3}\right) = 9 \cdot \tfrac{8}{27} - 2 + 4 = \tfrac{8}{3} + 2 = \tfrac{14}{3}$ .
$p(1) = 2 + a + b - 3 = 0 \Rightarrow a + b = 1$ . $p(-2) = -16 + 4a - 2b - 3 = -15 \Rightarrow 4a - 2b = 4 \Rightarrow 2a - b = 2$ . Adding: $3a = 3$ , so $a = 1$ , $b = 0$ .
$p(1) = 3 + k$ ; $p(-3) = -27 + 9k + 2 = 9k - 25$ . So $3 + k = 9k - 25$ , $8k = 28$ , $k = \tfrac{7}{2}$ .
$x^2 - x - 2 = (x - 2)(x + 1)$ . Remainder $Rx + S$ : $p(2) = 8 + 8 - 2 + 1 = 15 = 2R + S$ ; $p(-1) = -1 + 2 + 1 + 1 = 3 = -R + S$ . Subtracting: $3R = 12$ , $R = 4$ , $S = 7$ . Remainder $4x + 7$ .
(a) $p(1) = 1 + a + b$ and $p(-1) = -1 - a + b$ . Equal: $1 + a = -1 - a$ , so $a = -1$ . $p(2) = 8 - 2 + b = 0$ , so $b = -6$ . (b) $x^3 - x - 6 = (x - 2)(x^2 + 2x + 3)$ (compare coefficients: $x^2$ term $c - 2 = 0$ gives $c = 2$ ; constant $-2 \times 3 = -6$ ). Discriminant of $x^2 + 2x + 3$ is $4 - 12 = -8 < 0$ , so no real roots there. The only real root is $x = 2$ .
$p(x) \equiv (x - 1)(x - 3)q(x) + Rx + S$ with $R + S = 5$ and $3R + S = 1$ , so $R = -2$ , $S = 7$ : remainder $7 - 2x$ . Then $p(x) + 2x \equiv (x - 1)(x - 3)q(x) + 7$ , and $7$ has degree $0 < 2$ , so the remainder is $7$ .

Why one substitution is enough

Finding unknown coefficients

Combining a factor and a remainder

Remainders on dividing by a quadratic

Common mistakes

Summary

Practice

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