The Remainder Theorem and the Factor Theorem

The remainder without the division

In dividing polynomials, $2x^3 - 3x^2 + 4x - 5$ divided by $x - 2$ left a remainder of $7$. Now evaluate the polynomial at $2$:

$$p(2) = 2(8) - 3(4) + 4(2) - 5 = 16 - 12 + 8 - 5 = 7$$

The same $7$. That is not a coincidence, and it always happens:

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

Why the remainder is p(a)

Any division by $x - a$ can be written as

$$p(x) = (x - a)\,q(x) + r$$

where the remainder $r$ is a constant, because it has lower degree than $x - a$.

This equation holds for every $x$, so it holds at $x = a$:

$$p(a) = (a - a)\,q(a) + r = 0 \cdot q(a) + r = r$$

The factor $x - a$ becomes zero at $x = a$ and takes the whole quotient with it. What survives is exactly the remainder, and so $p(a) = r$.

The factor theorem

A factor divides evenly, which means a remainder of zero. Putting that together with the remainder theorem:

Factor theorem. $x - a$ is a factor of $p(x)$ if and only if $p(a) = 0$.

$$p(a) = 0 \iff (x - a) \text{ divides } p(x)$$

This connects three ideas that look different:

Statement	Means the same as
$p(a) = 0$	$a$ is a root of $p$
$x - a$ is a factor of $p(x)$	the division by $x - a$ has remainder $0$
the graph of $p$ crosses or touches the $x$-axis at $a$	$a$ is an $x$-intercept

A root, a factor and an $x$-intercept are one fact in three languages.

Testing for a factor

Is $x + 2$ a factor of $x^3 + 4x^2 + x - 6$?

$x + 2 = x - (-2)$, so evaluate at $-2$:

$$p(-2) = -8 + 16 - 2 - 6 = 0$$

The remainder is zero, so yes, $x + 2$ is a factor.

Is $x - 1$ a factor of $3x^3 + 2x^2 - x + 4$?

$$p(1) = 3 + 2 - 1 + 4 = 8 \ne 0$$

No. Dividing would leave a remainder of $8$.

Evaluating is usually faster than dividing, especially with a calculator — and it answers the yes-or-no question directly.

Factoring from one root

A cubic can be hard to factor by inspection. One root is enough to start.

Factor $x^3 - 2x^2 - 5x + 6$ completely.

1. Find a root. Try small integers that divide the constant $6$: $\pm 1$, $\pm 2$, $\pm 3$.

$$p(1) = 1 - 2 - 5 + 6 = 0$$

So $x - 1$ is a factor.

2. Divide it out. Synthetic division with $a = 1$ on $1, -2, -5, 6$ gives $1, -1, -6$ and remainder $0$:

$$x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6)$$

3. Factor the quadratic. $x^2 - x - 6 = (x - 3)(x + 2)$.

$$x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)$$

The roots are $1$, $3$ and $-2$. Each root found lowers the degree by one, so a cubic needs only one root before the quadratic methods take over.

Why try divisors of the constant? If the polynomial factors as $(x - r_1)(x - r_2)(x - r_3)$ with integer roots and leading coefficient $1$, the constant is $-r_1 r_2 r_3$, so every integer root divides it.

Worked examples

Common mistakes

Practice problems