The remainder theorem

Long divi­sion works for any divi­sor, but it is slow. When the divi­sor is lin­ear, such as $x - 3$ or $2x + 1$, there is a short­cut that finds the remain­der in one line, with­out divid­ing at all. It is called the remain­der the­o­rem, and it turns many divi­sion ques­tions into sim­ple sub­sti­tu­tion. In this les­son we see why it works and use it to find unknown coef­fi­cients.

The the­o­rem#

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

Why it is true. By the divi­sion algo­rithm, $f(x) = (x - a)\,q(x) + r$. The divi­sor $x - a$ has degree $1$, so the remain­der has degree $0$: it is just a num­ber $r$. This equa­tion is true for every value of $x$, so put $x = a$:

$$f(a) = (a - a)\,q(a) + r = 0 \times q(a) + r = r$$

So the remain­der $r$ equals $f(a)$.

A sim­ple way to remem­ber which num­ber to sub­sti­tute: set the divi­sor equal to zero and solve. For $x - a = 0$ we get $x = a$.

Divi­sors of the form ax − b and ax + b#

The same idea works when $x$ has a coef­fi­cient.

• When $f(x)$ is divided by $ax - b$, the remain­der is $\displaystyle f\left(\frac{b}{a}\right)$, because $ax - b = 0$ gives $\displaystyle x = \frac{b}{a}$.
• When $f(x)$ is divided by $ax + b$, the remain­der is $\displaystyle f\left(-\frac{b}{a}\right)$, because $ax + b = 0$ gives $\displaystyle x = -\frac{b}{a}$.

Exam­ple 1#

Find the remain­der when $f(x) = x^3 - 1$ is divided by $x + 1$.

$x + 1 = 0$ gives $x = -1$.

$$\text{Remainder} = f(-1) = (-1)^3 - 1 = -1 - 1 = -2$$

Exam­ple 2#

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

$$\begin{aligned} f(-2) &= 2(16) - 6(-8) + 2(4) - (-2) + 2 \\ &= 32 + 48 + 8 + 2 + 2 \\ &= 92 \end{aligned}$$

In the last les­son we found this same remain­der, $92$, after four rounds of long divi­sion. The the­o­rem gets it in one line. (It does not give the quo­tient, though; for that we still need to divide.)

Exam­ple 3#

Find the remain­der when $9x^3 - 3x^2 + x - 5$ is divided by $\displaystyle x - \frac{2}{3}$.

$$\displaystyle \begin{aligned} f\left(\tfrac{2}{3}\right) &= 9\left(\tfrac{8}{27}\right) - 3\left(\tfrac{4}{9}\right) + \tfrac{2}{3} - 5 \\ &= \tfrac{8}{3} - \tfrac{4}{3} + \tfrac{2}{3} - 5 \\ &= \tfrac{8 - 4 + 2}{3} - 5 \\ &= 2 - 5 = -3 \end{aligned}$$

The remain­der is $-3$.

Find­ing unknown coef­fi­cients#

When a ques­tion gives the remain­der and hides a coef­fi­cient, the the­o­rem gives an equa­tion for the unknown.

Exam­ple 4#

The poly­no­mi­als $ax^3 + 4x^2 + 3x - 4$ and $x^3 - 4x + a$ leave the same remain­der when divided by $x - 3$. Find $a$.

Let $f(x) = ax^3 + 4x^2 + 3x - 4$ and $g(x) = x^3 - 4x + a$. Both remain­ders come from $x = 3$, and they are equal, so $f(3) = g(3)$.

$$\begin{aligned} 27a + 36 + 9 - 4 &= 27 - 12 + a \\ 27a + 41 &= 15 + a \\ 26a &= 15 - 41 \\ 26a &= -26 \\ a &= -1 \end{aligned}$$

Exam­ple 5#

Find the qua­dratic poly­no­mial which, when divided by $x - 1$, $x - 2$ and $x - 3$, leaves remain­ders $1$, $2$ and $4$ respec­tively.

Let $f(x) = ax^2 + bx + c$. By the remain­der the­o­rem the three remain­ders are $f(1)$, $f(2)$ and $f(3)$.

$$f(1) = 1 \implies a + b + c = 1 \quad (1)$$

$$f(2) = 2 \implies 4a + 2b + c = 2 \quad (2)$$

$$f(3) = 4 \implies 9a + 3b + c = 4 \quad (3)$$

Sub­tract to remove $c$:

$$(2) - (1): \quad 3a + b = 1$$

$$(3) - (2): \quad 5a + b = 2$$

Sub­tract­ing these two gives $2a = 1$, so $\displaystyle a = \frac{1}{2}$. Then $\displaystyle b = 1 - 3a = 1 - \frac{3}{2} = -\frac{1}{2}$, and from (1), $c = 1 - a - b = 1$.

$$\displaystyle f(x) = \frac{1}{2}x^2 - \frac{1}{2}x + 1$$

Check: $\displaystyle f(3) = \frac{9}{2} - \frac{3}{2} + 1 = 3 + 1 = 4$. Cor­rect.

Prac­tice#

1. Find the remain­der when $x^4 - 3x^2 + 2x + 1$ is divided by $x - 1$.
2. Find the remain­der when $x^3 + 4x^2 - 3x + 10$ is divided by $x + 4$.
3. Find the remain­der when $4x^3 - 12x^2 + 14x - 3$ is divided by $2x - 1$.
4. Find the remain­der when $x^3 - 6x^2 + 2x - 4$ is divided by $3x - 1$.
5. Find the remain­der when $x^3 + 3x^2 + 3x + 1$ is divided by (a) $x + 1$ (b) $\displaystyle x - \frac{1}{2}$ (c) $x$ (d) $5 + 2x$.
6. When $kx^3 + 9x^2 + 4x - 8$ is divided by $x + 3$, the remain­der is $-20$. Find $k$.
7. The poly­no­mi­als $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remain­der when divided by $x - 2$. Find $a$.
8. When $x^2 + bx + c$ is divided by $x - 1$ the remain­der is $0$, and when it is divided by $x + 2$ the remain­der is $12$. Find $b$ and $c$.
9. When $2x^3 + 3x^2 + ax + b$ is divided by $x - 2$ the remain­der is $2$, and when it is divided by $x + 2$ the remain­der is $-2$. Find $a$ and $b$.
10. When $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$ is divided by $x - 1$ and $x + 1$, the remain­ders are $5$ and $19$. Find the remain­der when $f(x)$ is divided by $x - 2$.

Answers#