Remainder Theorem — Statement, Proof & Examples

The Remainder Theorem lets you find a polynomial remainder without doing long division. If you divide $P(x)$ by $x-a$, the remainder is $P(a)$.

This only works when the divisor is written in the form $x-a$. For $x-3$, use $a=3$. For $x+2$, use $a=-2$.

Remainder Theorem Statement

If a polynomial $P(x)$ is divided by $x-a$, then

$$\text{remainder} = P(a)$$

This is the full idea of the theorem. A division question becomes a substitution question.

Why The Remainder Is $P(a)$

When you divide a polynomial $P(x)$ by a linear expression $x-a$, the division algorithm says

$$P(x) = (x-a)Q(x) + r$$

where $Q(x)$ is the quotient and $r$ is the remainder. Because the divisor has degree $1$, the remainder must have degree less than $1$, so $r$ is just a constant.

Now substitute $x=a$:

$$P(a) = (a-a)Q(a) + r = 0 + r = r$$

So the remainder is $P(a)$.

Worked Example: Divide By $x-2$

Find the remainder when

$$P(x) = x^3 + 2x^2 - 5x + 1$$

is divided by $x-2$.

Because the divisor is $x-2$, use $a=2$. Then evaluate $P(2)$:

$$P(2) = 2^3 + 2(2^2) - 5(2) + 1$$ $$= 8 + 8 - 10 + 1 = 7$$

So the remainder is

$$7$$

You do not need the quotient to answer this question. Once you have $P(2)$, you already have the remainder.

How To Use The Remainder Theorem

For most problems, the process is short:

1. Rewrite the divisor as $x-a$.
2. Identify $a$ correctly.
3. Compute $P(a)$.
4. Report that value as the remainder.

If $P(a)=0$, the remainder is zero, which means $x-a$ divides the polynomial exactly.

How It Connects To The Factor Theorem

The Factor Theorem is a direct consequence of the Remainder Theorem.

If

$$P(a)=0$$

then the remainder on division by $x-a$ is $0$, so $x-a$ is a factor of $P(x)$.

So the Remainder Theorem tells you the remainder in every case, and the Factor Theorem focuses on the special case where the remainder is zero.

Common Mistakes Students Make

Using The Wrong Sign For $a$

For $x-4$, use $a=4$. For $x+4$, use $a=-4$. This is the most common error.

Forgetting The Divisor Must Match $x-a$

The theorem is stated for divisors of the form $x-a$. If the divisor is $2x-3$, you cannot plug in $3$ and call that the remainder.

For a divisor like $2x-3$, set $2x-3=0$ first, so $x=\frac{3}{2}$. Then the remainder is $P\left(\frac{3}{2}\right)$ because the remainder is still a constant when dividing by a linear polynomial.

Mixing Up Quotient And Remainder

$P(a)$ gives the remainder only. It does not give the quotient.

When The Remainder Theorem Is Useful

You will usually see it when you want to:

• find a polynomial remainder quickly
• check whether a linear expression might be a factor
• connect a substitution value to synthetic division
• avoid full polynomial long division in a simple case

Try A Similar Problem

Take

$$P(x) = 2x^3 - 3x + 5$$

and find the remainder when dividing by $x+1$. Start by rewriting the divisor as $x-(-1)$, so you know to compute $P(-1)$. If you want a good check, compare your answer with synthetic division and make sure the remainder matches.