Remainder Theorem — Statement, Proof & Examples

The Remainder Theorem turns a polynomial division into a single substitution: if you divide $P(x)$ by $x-a$, the remainder is just $P(a)$. No long division required.

The formula and its symbols

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

Here $P(x)$ is the polynomial being divided, $x-a$ is the linear divisor, and $a$ is the number that makes the divisor zero. The theorem applies only when the divisor is in the form $x-a$: for $x-3$, use $a=3$; for $x+2$, use $a=-2$. A division question becomes a substitution question.

Why the remainder is $P(a)$

This is the part worth understanding, not just memorizing. When you divide $P(x)$ by a linear expression $x-a$, the division algorithm gives

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 a constant. Now substitute $x=a$:

The $(a-a)$ factor kills the quotient term, leaving exactly $r$. So the remainder is $P(a)$.

Worked example: divide by $x-2$

Find the remainder when

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

So the remainder is $7$. You never needed the quotient — once you have $P(2)$, you have the remainder.

Practice the calculation

Find the remainder when

is divided by $x+1$. First rewrite the divisor as $x-(-1)$, so you know to compute $P(-1)$. Answer check: $P(-1) = 2(-1)^3 - 3(-1) + 5 = -2 + 3 + 5 = 6$, so the remainder is $6$. If you want a second confirmation, run synthetic division and verify the remainder matches.

Calculation traps to watch for

Using the wrong sign for $a$

For $x-4$, use $a=4$; for $x+4$, use $a=-4$. This sign slip is the most common error. Self-check: the value of $a$ is always the root of the divisor, the number that makes $x-a$ equal to zero.

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. Instead set $2x-3=0$ first, so $x=\frac{3}{2}$, then the remainder is $P\left(\frac{3}{2}\right)$ — still a constant, because dividing by any linear polynomial leaves a constant remainder.

Mixing up quotient and remainder

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

How it connects to the Factor Theorem

The Factor Theorem is a direct consequence. If

then the remainder on division by $x-a$ is $0$, so $x-a$ is a factor of $P(x)$. The Remainder Theorem tells you the remainder in every case; the Factor Theorem zooms in on the special case where that remainder is zero.

When the Remainder Theorem is useful

You will reach for it to find a polynomial remainder quickly, check whether a linear expression might be a factor, connect a substitution value to synthetic division, or avoid full polynomial long division in a simple case.