The factor theorem

When does the remainder vanish?

From the remainder theorem, when $p(x)$ is divided by $x - a$, the remainder is $p(a)$. If $p(a) = 0$, then there is no remainder, and $x - a$ divides $p(x)$ exactly. This is the factor theorem.

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

"If and only if" means that it works both ways. If $p(a) = 0$, then $x - a$ is a factor. If $x - a$ is a factor, then $p(a) = 0$.

Think of $12$ and $4$. Because $12 \div 4$ has no remainder, $4$ is a factor of $12$. The factor theorem is the same idea for polynomials, and it lets us test a factor by putting in a number, instead of doing a long division.

Using the factor theorem

Example 1. Is $x + 1$ a factor of $x^3 + x^2 + x + 1$?

$x + 1 = x - (-1)$, so $a = -1$. $p(-1) = -1 + 1 - 1 + 1 = 0$. Yes, it is a factor.

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

$p(1) = 1 + 1 + 1 = 3$. This is not $0$, so no.

Example 3. Find $k$ if $x - 1$ is a factor of $kx^2 - 3x + 2$.

$p(1) = k - 3 + 2 = 0$, so $k = 1$.

Example 4. A divisor with a number in front of x. Is $2x - 1$ a factor of $2x^3 - x^2 + 4x - 2$? The zero of $2x - 1$ is $\dfrac{1}{2}$. $p\left(\dfrac{1}{2}\right) = \dfrac{2}{8} - \dfrac{1}{4} + 2 - 2 = 0$. Yes.

Finding the other factor

Example 5. Show that $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6$, and find the other factor.

$p(-2) = -8 + 12 - 10 + 6 = 0$. So $x + 2$ is a factor.

By synthetic division with $a = -2$: the coefficients $1, 3, 5, 6$ give $1,\ 3 - 2 = 1,\ 5 - 2 = 3,\ 6 - 6 = 0$. The remainder is $0$, and the quotient is $x^2 + x + 3$.

Example 6. All the factors of a cubic. Factorise $x^3 - 4x^2 + x + 6$.

• Try $x = 3$: $27 - 36 + 3 + 6 = 0$. So $x - 3$ is a factor.
• Synthetic division by $3$: $1,\ -4 + 3 = -1,\ 1 - 3 = -2,\ 6 - 6 = 0$. The quotient is $x^2 - x - 2$.
• Now $x^2 - x - 2 = (x - 2)(x + 1)$.

So $x^3 - 4x^2 + x + 6 = (x - 3)(x - 2)(x + 1)$. The zeroes are $3$, $2$ and $-1$, and their sum is $4$, the negative of the coefficient of $x^2$. That is a good check.

Whatever zeroes you choose, each one divides the constant term. That is why we test only the divisors of the constant term.

Which numbers to try?

For a polynomial with whole-number coefficients, a whole-number zero must divide the constant term. For $x^3 - 6x^2 + 11x - 6$, the constant term is $-6$. So we try $\pm 1, \pm 2, \pm 3, \pm 6$.

Why? If $a$ is a zero and $p(x) = (x - a)q(x)$, then the constant term of $p$ is $-a$ times the constant term of $q$. So $a$ divides the constant term.

Practise with new numbers

Try it in Python

This program tries every divisor of the constant term, with both signs, and prints the factors.

Common mistakes

• Using the wrong sign. For $x + 2$, test $p(-2)$. Not $p(2)$.
• Stopping after one factor. A cubic has three factors, and the quotient may factorise more.
• Trying numbers that do not divide the constant term. They cannot be whole-number zeroes.
• Forgetting to check the remainder in synthetic division. It must be $0$.

Remember these

• $x - a$ is a factor of $p(x)$ if and only if $p(a) = 0$.
• To test $x + 2$, use $a = -2$.
• A whole-number zero divides the constant term. We try these numbers first.
• After one factor, divide, and factorise the quotient.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $k$ if $x + 1$ is a factor of $x^2 + kx + 2$.

2. Which numbers should we try as zeroes of $x^3 + 2x^2 - x - 2$?

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

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