The Remainder Theorem

Dividing polynomials takes a few steps. But if all you need is the remainder, there’s a shortcut: substitute one number into the polynomial. This is the remainder theorem. It saves time, and it’s the key step towards the factor theorem and factoring cubics and quartics.

Key ideas

The remainder theorem

When a polynomial $P(x)$ is divided by $x - a$, the remainder is $P(a)$.

Here’s why. Any division by $x - a$ can be written as

$$P(x) = (x - a)\,Q(x) + R$$

This is true for every $x$, so put in $x = a$. The first term becomes $(a - a)Q(a) = 0 \cdot Q(a) = 0$, which leaves

$$P(a) = R$$

For example, dividing $P(x) = x^3 + 4x^2 - 3x + 5$ by $x - 1$ gives remainder $7$ (that’s Example 1 on the division page). Check: $P(1) = 1 + 4 - 3 + 5 = 7$. ✓

Picking the right value to substitute

Substitute the value of $x$ that makes the divisor zero:

Divisor	Set it to zero	Remainder
$x - 3$	$x = 3$	$P(3)$
$x + 2$	$x = -2$	$P(-2)$
$2x - 1$	$x = \tfrac{1}{2}$	$P\left(\tfrac{1}{2}\right)$
$ax - b$	$x = \tfrac{b}{a}$	$P\left(\tfrac{b}{a}\right)$

The last row works for the same reason: $P(x) = (ax - b)Q(x) + R$, and $ax - b = 0$ when $x = \tfrac{b}{a}$.

What it looks like on a graph

$P(a)$ is the $y$-coordinate of the point on the graph of $y = P(x)$ where $x = a$. So the remainder when you divide by $x - a$ is the height of the graph at $x = a$.

For $P(x) = x^3 - 3x^2 - x + 4$, the points $(-1, 1)$ and $(2, -2)$ give the remainders on dividing by $x + 1$ and by $x - 2$.

Finding unknown coefficients

If a polynomial has an unknown coefficient, like $k$, and you know a remainder, the remainder theorem gives you an equation in $k$. Two unknowns need two conditions, which give a system of two equations.

Worked examples

Example 1: Two remainders

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

Solution. (a) Substitute $x = 2$:

$$P(2) = 2^3 - 3(2)^2 - 2 + 4 = 8 - 12 - 2 + 4 = -2$$

The remainder is $-2$.

(b) The divisor $x + 1$ is zero when $x = -1$:

$$P(-1) = (-1)^3 - 3(-1)^2 - (-1) + 4 = -1 - 3 + 1 + 4 = 1$$

The remainder is $1$. These are the two points marked on the graph above.

Check (a) with synthetic division:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 1 & -3 & -1 & 4 \\ & & 2 & -2 & -6 \\ \hline & 1 & -1 & -3 & \boxed{-2} \end{array}$$

Example 2: A divisor of the form ax − b

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

Solution. $2x - 1 = 0$ when $x = \tfrac{1}{2}$:

$$\begin{aligned} P\left(\tfrac{1}{2}\right) &= 4\left(\tfrac{1}{8}\right) + 2\left(\tfrac{1}{4}\right) - 5\left(\tfrac{1}{2}\right) + 1 \\ &= \tfrac{1}{2} + \tfrac{1}{2} - \tfrac{5}{2} + 1 \\ &= -\tfrac{1}{2} \end{aligned}$$

The remainder is $-\tfrac{1}{2}$.

Example 3: Finding k from a remainder

When $x^3 + kx^2 - 4x + 3$ is divided by $x - 2$, the remainder is $7$. Find $k$.

Solution. By the remainder theorem, $P(2) = 7$:

$$\begin{aligned} 2^3 + k(2)^2 - 4(2) + 3 &= 7 \\ 8 + 4k - 8 + 3 &= 7 \\ 4k &= 4 \\ k &= 1 \end{aligned}$$

Check: with $k = 1$, $P(2) = 8 + 4 - 8 + 3 = 7$. ✓

Example 4: The same remainder for two divisors

Find $k$ so that $f(x) = x^3 + 2x^2 + kx + 5$ leaves the same remainder when divided by $x - 1$ and by $x + 3$.

Solution. The two remainders are $f(1)$ and $f(-3)$:

$$f(1) = 1 + 2 + k + 5 = 8 + k$$ $$f(-3) = -27 + 18 - 3k + 5 = -4 - 3k$$

Set them equal:

$$8 + k = -4 - 3k \quad\Rightarrow\quad 4k = -12 \quad\Rightarrow\quad k = -3$$

Check: with $k = -3$, $f(1) = 8 - 3 = 5$ and $f(-3) = -4 + 9 = 5$. Both remainders are $5$. ✓

Common mistakes

Substituting the wrong sign. For the divisor $x + 1$, the remainder is $P(-1)$, not $P(1)$. Always ask: what value of $x$ makes the divisor zero?

Using the coefficient instead of the zero for ax − b. For $2x - 1$, substitute $x = \tfrac{1}{2}$, not $x = 1$ or $x = 2$.

Making arithmetic errors with negatives and powers. $(-1)^2 = 1$, so $-3(-1)^2 = -3$, but $(-1)^3 = -1$. Put every substituted value in brackets and work slowly.

Thinking the remainder theorem gives the quotient. $P(a)$ gives only the remainder. If you also need the quotient, you still have to divide.

Solving for k and stopping. After finding $k$, substitute it back and check the remainder (or both remainders) comes out right. It takes ten seconds and catches most slips.

Practice

1. (Warm-up) Find the remainder when $x^3 + 2x - 7$ is divided by $x - 1$.

Solution

$P(1) = 1 + 2 - 7 = -4$. The remainder is $-4$.

2. (Warm-up) Which value would you substitute to find the remainder on dividing by each divisor?

• (a) $x - 4$
• (b) $x + 5$
• (c) $2x - 3$
• (d) $3x + 1$

Solution

(a) $x = 4$

(b) $x = -5$

(c) $x = \tfrac{3}{2}$

(d) $x = -\tfrac{1}{3}$

3. (Warm-up) Find the remainder when $2x^3 - x^2 + 3$ is divided by $x + 2$. Check with synthetic division.

Solution

$$P(-2) = 2(-8) - 4 + 3 = -16 - 4 + 3 = -17$$

Synthetic division with $a = -2$ (and $0$ for the missing $x$ term):

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} -2 & 2 & -1 & 0 & 3 \\ & & -4 & 10 & -20 \\ \hline & 2 & -5 & 10 & \boxed{-17} \end{array}$$

Both give a remainder of $-17$.

4. (Core) Find the remainder when $x^4 - 2x^3 + 5x - 6$ is divided by $x - 3$.

Solution

$$P(3) = 81 - 2(27) + 15 - 6 = 81 - 54 + 15 - 6 = 36$$

The remainder is $36$.

5. (Core) Find the remainder when $6x^3 - x^2 + 4x + 2$ is divided by $2x + 1$.

Solution

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

$$\begin{aligned} P\left(-\tfrac{1}{2}\right) &= 6\left(-\tfrac{1}{8}\right) - \tfrac{1}{4} + 4\left(-\tfrac{1}{2}\right) + 2 \\ &= -\tfrac{3}{4} - \tfrac{1}{4} - 2 + 2 \\ &= -1 \end{aligned}$$

The remainder is $-1$.

6. (Core) When $2x^3 + kx^2 - x + 6$ is divided by $x + 1$, the remainder is $9$. Find $k$.

Solution

$$P(-1) = -2 + k + 1 + 6 = 5 + k$$

Set $5 + k = 9$, so $k = 4$.

Check: $P(-1) = -2 + 4 + 1 + 6 = 9$. ✓

7. (Core) Find $k$ so that $x^3 - kx^2 + 2x + 1$ leaves the same remainder when divided by $x - 2$ and by $x + 1$. What is that remainder?

Solution

$$P(2) = 8 - 4k + 4 + 1 = 13 - 4k$$$$P(-1) = -1 - k - 2 + 1 = -2 - k$$

Set them equal: $13 - 4k = -2 - k$, so $15 = 3k$ and $k = 5$.

The remainder is $13 - 4(5) = -7$. Check: $-2 - 5 = -7$. ✓

8. (Challenge) When $P(x) = x^3 + ax^2 + bx - 6$ is divided by $x - 1$, the remainder is $-4$. When it is divided by $x - 2$, the remainder is $8$. Find $a$ and $b$.

Solution

$P(1) = -4$:

$$1 + a + b - 6 = -4 \quad\Rightarrow\quad a + b = 1$$

$P(2) = 8$:

$$8 + 4a + 2b - 6 = 8 \quad\Rightarrow\quad 4a + 2b = 6 \quad\Rightarrow\quad 2a + b = 3$$

Subtract the first equation from the second: $a = 2$. Then $b = 1 - 2 = -1$.

Check: $P(x) = x^3 + 2x^2 - x - 6$, so $P(1) = 1 + 2 - 1 - 6 = -4$ and $P(2) = 8 + 8 - 2 - 6 = 8$. ✓

9. (Challenge) When $P(x)$ is divided by $x - 1$ the remainder is $3$, and when it is divided by $x + 2$ the remainder is $-3$. Find the remainder when $P(x)$ is divided by $(x - 1)(x + 2)$.

Solution

The divisor $(x - 1)(x + 2)$ is quadratic, so the remainder has degree at most $1$. Call it $mx + c$:

$$P(x) = (x - 1)(x + 2)\,Q(x) + mx + c$$

Substitute $x = 1$: the first term is $0$, so $P(1) = m + c = 3$.

Substitute $x = -2$: $P(-2) = -2m + c = -3$.

Subtracting the second equation from the first gives $3m = 6$, so $m = 2$ and $c = 1$.

The remainder is $2x + 1$.