Factor Theorem — How to Factorize Polynomials

The factor theorem says that for a polynomial $P(x)$, the linear factor $(x-a)$ works exactly when $P(a)=0$. That makes it one of the fastest ways to check whether a guessed root actually helps you factor a polynomial.

$$P(a)=0 \iff (x-a) \text{ is a factor of } P(x).$$

Instead of doing full division for every guess, you substitute one value. If the result is zero, you have found both a root, $a$, and a matching linear factor, $(x-a)$.

What The Factor Theorem Means

If $(x-a)$ is a factor, then the polynomial can be written as

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

for some polynomial $Q(x)$. Now plug in $x=a$:

$$P(a)=(a-a)Q(a)=0.$$

So a factor creates a zero. The reverse matters just as much: if substituting $a$ gives zero, then division by $(x-a)$ leaves remainder $0$, so $(x-a)$ really is a factor.

This is why the theorem connects roots and factors. If $P(a)=0$, then $a$ is a root of the polynomial, and $(x-a)$ is the corresponding linear factor.

How To Use The Factor Theorem To Factor Polynomials

The theorem does not tell you which value of $a$ to try first. It tells you how to test a candidate once you have one.

In many school problems, sensible integer candidates come from the factors of the constant term. For a monic polynomial, values like $\pm 1$, $\pm 2$, and other divisors of the constant term are often worth checking first. That is a strategy, not a guarantee.

The usual workflow is short:

1. Choose a candidate value $a$.
2. Compute $P(a)$.
3. If $P(a)=0$, write down $(x-a)$ as a factor.
4. Divide the polynomial by $(x-a)$ and keep factoring if possible.

Worked Example: Factor A Cubic

Factor

$$P(x)=x^3-6x^2+11x-6.$$

A sensible first test is $x=1$:

$$P(1)=1-6+11-6=0.$$

Because $P(1)=0$, the factor theorem tells us that $(x-1)$ is a factor. Now divide by $(x-1)$ to get the quotient:

$$x^2-5x+6.$$

So

$$x^3-6x^2+11x-6=(x-1)(x^2-5x+6).$$

The quadratic factors further:

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

That gives the full factorization:

$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3).$$

The key move was not magic. We tested one value, found a zero, turned that zero into a factor, and then finished with ordinary factoring.

Common Factor Theorem Mistakes

Mixing Up The Sign

If $P(2)=0$, the factor is $(x-2)$, not $(x+2)$.

The theorem matches the form $(x-a)$, so the sign in the factor is the opposite of the sign inside the value $a$.

Stopping Too Early

Finding one factor is often only the first step. After you get $(x-a)$, divide the polynomial and factor the quotient if possible.

Assuming Every Polynomial Has Easy Integer Tests

Testing small integers is useful only when the problem structure supports it. Some polynomials have rational, irrational, or complex roots instead of simple integer roots.

Forgetting The Condition

The theorem applies to polynomials. It is not a general shortcut for every algebraic expression.

When The Factor Theorem Is Useful

The factor theorem is especially useful when you need to:

• test whether a guessed root actually works
• factor a polynomial step by step
• connect zeros to linear factors
• set up synthetic division more efficiently

It is often used together with the remainder theorem and synthetic division. In fact, when you divide by $(x-a)$, the remainder is $P(a)$. The factor theorem is the special case where that remainder is $0$.

Try A Similar Problem

Try the same process on

$$x^3-4x^2-x+4.$$

Test a simple candidate value first, use the factor theorem to confirm a linear factor, and then finish the factorization by division. As a quick check, expand your final factors to make sure you recover the original polynomial.