The factor theorem

When you divide a polynomial $p(x)$ by $x - a$ , the division leaves a remainder, and that remainder is the value of the polynomial at $x = a$ :

p(x) = (x - a)Q(x) + R, \qquad R = p(a).

That is the remainder theorem. If the remainder is zero, then $x - a$ divides the polynomial exactly, with nothing left over:

x - a \text{ is a factor of } p(x) \quad \Longleftrightarrow \quad p(a) = 0.

That is the factor theorem. This page proves the first from the division, and reads the second from it.

Before you start

You need three facts.

Algebraic division. Dividing $p(x)$ by $x - a$ gives a quotient $Q(x)$ and a remainder $R$ , with $p(x) = (x - a)Q(x) + R$ .
Degree of the remainder. The remainder has a lower degree than the divisor. The divisor $x - a$ has degree $1$ , so the remainder has degree $0$ — it is a constant, not an expression in $x$ .
Substitution. To work out $p(a)$ , replace every $x$ by the number $a$ and work out the value.

Predict first

Take $p(x) = x^3 - x - 6$ .

Before you read on, predict: is $x - 2$ a factor of $p(x)$ ? Work out $p(2)$ by hand, and write down whether you expect the division by $x - 2$ to come out exactly. Then read the proof and see what that substitution is really computing.

The remainder theorem

When you divide $p(x)$ by $x - a$ , do the division as far as it goes. You get a quotient and a remainder:

p(x) = (x - a)Q(x) + R.

The divisor $x - a$ has degree $1$ . The remainder must have a lower degree, so $R$ is a constant. It is a number, not an expression in $x$ .

Now substitute $x = a$ into both sides. This is allowed: the two sides are equal for every value of $x$ , so they are equal at $x = a$ .

p(a) = (a - a)Q(a) + R.

The first bracket is zero, so the first term on the right disappears:

p(a) = 0 \times Q(a) + R = R.

So the remainder is exactly $p(a)$ . Multiplying by $Q(a)$ is not needed to find it, and neither is the division. Substituting one number gives the remainder.

What a zero remainder means

A factor divides exactly. That means the division leaves no remainder, so $R = 0$ . Put $R = 0$ into the division identity:

p(x) = (x - a)Q(x).

The right-hand side is $x - a$ multiplied by a whole polynomial, and it equals $p(x)$ . So $x - a$ is a factor of $p(x)$ .

The remainder theorem says $R = p(a)$ . So $R = 0$ is the same statement as $p(a) = 0$ . Putting the two together gives the factor theorem:

x - a \text{ is a factor of } p(x) \quad \Longleftrightarrow \quad p(a) = 0.

The double arrow means both directions hold. If $x - a$ is a factor, then the division has no remainder and $p(a) = 0$ . If $p(a) = 0$ , then the remainder is zero, so the division has nothing left over and $x - a$ is a factor. Each direction is the same division identity read a different way.

Why this is useful. Finding a factor of a cubic takes work; substituting a number takes one line. The factor theorem turns the division into a substitution. It also explains why one root is enough to start factorising: if $p(a) = 0$ , then $p(x) = (x - a)Q(x)$ , and $Q$ has degree one less than $p$ . A cubic becomes a quadratic, which you can factorise in the usual way.

A factor with a coefficient: (ax - b)

The theorem is written for $x - a$ . A linear factor may have a number in front of the $x$ , such as $ax - b$ . Take out the coefficient first:

ax - b = a\left(x - \frac{b}{a}\right).

The constant $a$ does not change whether the division is exact. Multiplying a polynomial by a non-zero number cannot turn a zero remainder into a non-zero one, or the other way round. So $ax - b$ is a factor of $p(x)$ exactly when $x - \frac{b}{a}$ is a factor, which is exactly when

p\left(\frac{b}{a}\right) = 0.

Example. Is $2x - 1$ a factor of $p(x) = 2x^3 - x^2 + 2x - 1$ ? Here the value to substitute is $x = \frac{1}{2}$ :

p\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{8} - \frac{1}{4} + 2 \cdot \frac{1}{2} - 1 = \frac{1}{4} - \frac{1}{4} + 1 - 1 = 0.

The value is zero, so $2x - 1$ is a factor. Indeed, $2x^3 - x^2 + 2x - 1 = (2x - 1)(x^2 + 1)$ .

Check your understanding

Work out the remainder when $x^3 - 4x^2 + 2x + 5$ is divided by $x - 3$ .
Show that $x + 1$ is a factor of $p(x) = x^3 + 2x^2 - 5x - 6$ .
Is $3x + 1$ a factor of $p(x) = 3x^3 + x^2 + 3x + 1$ ? Give a reason.

Answers

By the remainder theorem the remainder is $p(3)$ : $p(3) = 27 - 36 + 6 + 5 = 2$ . The remainder is $2$ .
Write $x + 1$ as $x - (-1)$ , so $a = -1$ . Then $p(-1) = -1 + 2 + 5 - 6 = 0$ . The value is zero, so by the factor theorem $x + 1$ is a factor.
Write $3x + 1$ as $3\left(x + \frac{1}{3}\right)$ , so the value to substitute is $x = -\frac{1}{3}$ . Then $p\left(-\frac{1}{3}\right) = 3 \cdot \left(-\frac{1}{27}\right)
\frac{1}{9} - 1 + 1 = -\frac{1}{9} + \frac{1}{9} = 0$. The value is zero, so $3x + 1$ is a factor.

A common error

For the divisor $x - a$ , substitute $x = a$ , with the sign that makes the bracket zero. The divisor $x + 3$ is $x - (-3)$ , so substitute $x = -3$ , not $x = 3$ . Substituting the wrong sign turns a factor into a non-factor, or the reverse.

The other trap is treating a non-zero remainder as if it were zero. If $p(2) = 7$ , then $x - 2$ is not a factor, and the division is not finished. The full statement is

p(x) = (x - 2)Q(x) + 7.

The value $7$ is the remainder, and it belongs in the answer.