The Factor Theorem

The remainder theorem says that dividing $P(x)$ by $x - a$ leaves remainder $P(a)$. So what happens when $P(a) = 0$? There’s no remainder, which means $x - a$ divides evenly: it’s a factor. This is the factor theorem, and it’s how you factor cubics and quartics that don’t fit any of the familiar patterns.

Key ideas

The factor theorem

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

“If and only if” means it works both ways:

• If $P(a) = 0$, the remainder on dividing by $x - a$ is $0$, so $P(x) = (x - a)Q(x)$ and $x - a$ is a factor.
• If $x - a$ is a factor, then $P(x) = (x - a)Q(x)$, so $P(a) = 0 \cdot Q(a) = 0$.

The same idea works for $ax - b$: it’s a factor if and only if $P\left(\tfrac{b}{a}\right) = 0$.

Zeros, factors and x-intercepts

These three statements say the same thing:

Zero	Factor	Graph
$P(a) = 0$	$x - a$ is a factor	the graph crosses or touches the $x$-axis at $(a, 0)$

The zeros of $x^3 - 2x^2 - 5x + 6$ are $-2$, $1$ and $3$, so its factors are $x + 2$, $x - 1$ and $x - 3$.

Which values should you test?

You can’t test every number, so narrow it down.

Integer zeros. If $P(x)$ has integer coefficients and $x - b$ is a factor (with $b$ an integer), then $b$ is a factor of the constant term. For $x^3 - 2x^2 - 5x + 6$, the only integers worth testing are the factors of $6$: $\pm 1, \pm 2, \pm 3, \pm 6$.

Why? If $P(x) = (x - b)(\dots)$, then multiplying the constant terms of the factors gives the constant term of $P(x)$. So $b$ must divide it.

Rational zeros. If the leading coefficient isn’t $1$, a zero might be a fraction. Any rational zero $\tfrac{b}{a}$ (in lowest terms) has $b$ a factor of the constant term and $a$ a factor of the leading coefficient. For $2x^3 + x^2 + 5x - 3$, the candidates are

$$\pm 1,\ \pm 3,\ \pm\tfrac{1}{2},\ \pm\tfrac{3}{2}$$

Test the integers first, since they’re easiest.

Factoring a cubic or quartic

1. List the possible zeros and test them until you find a value $a$ with $P(a) = 0$.
2. Divide $P(x)$ by $x - a$ (synthetic division is quickest). The remainder should be $0$.
3. Factor the quotient. If it’s a quadratic, factor it the usual way. If it’s a cubic, repeat steps 1 and 2.
4. Write $P(x)$ as a product of all the factors, and check by expanding or by substituting a value.

Sometimes the last quadratic doesn’t factor. If its discriminant $b^2 - 4ac$ is negative, it has no real zeros, so it can’t be factored any further using real numbers.

Worked examples

Example 1: Is it a factor?

Let $P(x) = x^3 - 3x^2 - 4x + 12$. Is $x - 2$ a factor? Is $x + 1$ a factor?

Solution. Test $x = 2$:

$$P(2) = 8 - 12 - 8 + 12 = 0$$

So $x - 2$ is a factor.

For $x + 1$, test $x = -1$:

$$P(-1) = -1 - 3 + 4 + 12 = 12$$

$P(-1) \ne 0$, so $x + 1$ is not a factor. (Dividing by $x + 1$ would leave remainder $12$.)

Example 2: Factoring a cubic

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

Solution. The possible integer zeros are the factors of $6$: $\pm 1, \pm 2, \pm 3, \pm 6$. Try $x = 1$:

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

So $x - 1$ is a factor. Divide using synthetic division:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 1 & -2 & -5 & 6 \\ & & 1 & -1 & -6 \\ \hline & 1 & -1 & -6 & \boxed{0} \end{array}$$

The quotient is $x^2 - x - 6 = (x - 3)(x + 2)$. So

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

Check with the graph above: the $x$-intercepts are $1$, $3$ and $-2$. ✓

Example 3: A rational zero

Factor $P(x) = 2x^3 + x^2 + 5x - 3$ as fully as possible.

Solution. The possible rational zeros are $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$. The integers don’t work: $P(1) = 5$, $P(-1) = -9$, $P(3) = 75$ and $P(-3) = -63$. Try $x = \tfrac{1}{2}$:

$$P\left(\tfrac{1}{2}\right) = 2\left(\tfrac{1}{8}\right) + \tfrac{1}{4} + \tfrac{5}{2} - 3 = \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{5}{2} - 3 = 0$$

So $2x - 1$ is a factor. Divide $P(x)$ by $2x - 1$ (long division):

• $2x^3 \div 2x = x^2$. Subtract $x^2(2x - 1) = 2x^3 - x^2$ to get $2x^2 + 5x$.
• $2x^2 \div 2x = x$. Subtract $x(2x - 1) = 2x^2 - x$ to get $6x - 3$.
• $6x \div 2x = 3$. Subtract $3(2x - 1) = 6x - 3$ to get $0$.

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

For $x^2 + x + 3$, the discriminant is $1^2 - 4(1)(3) = -11 \lt 0$, so it doesn’t factor further. That’s the full factorization.

Example 4: Factoring a quartic

Factor $P(x) = x^4 + 2x^3 - 7x^2 - 8x + 12$.

Solution. The possible integer zeros are the factors of $12$. Try $x = 1$: $P(1) = 1 + 2 - 7 - 8 + 12 = 0$. Divide by $x - 1$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrrr} 1 & 1 & 2 & -7 & -8 & 12 \\ & & 1 & 3 & -4 & -12 \\ \hline & 1 & 3 & -4 & -12 & \boxed{0} \end{array}$$

The quotient is the cubic $x^3 + 3x^2 - 4x - 12$. Repeat: try $x = 2$, which gives $8 + 12 - 8 - 12 = 0$. Divide by $x - 2$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 1 & 3 & -4 & -12 \\ & & 2 & 10 & 12 \\ \hline & 1 & 5 & 6 & \boxed{0} \end{array}$$

The quotient is $x^2 + 5x + 6 = (x + 2)(x + 3)$. So

$$x^4 + 2x^3 - 7x^2 - 8x + 12 = (x - 1)(x - 2)(x + 2)(x + 3)$$

Check with $x = 0$: the left side is $12$, and the right side is $(-1)(-2)(2)(3) = 12$. ✓

(You could also factor the cubic $x^3 + 3x^2 - 4x - 12$ by grouping: $x^2(x + 3) - 4(x + 3)$. See factoring polynomials.)

Common mistakes

Mixing up the zero and the factor. If $P(-3) = 0$, the factor is $x + 3$, not $x - 3$. The factor is ”$x$ minus the zero”.

Testing values that can’t work. For integer coefficients, only factors of the constant term can be integer zeros. Testing $x = 5$ for $x^3 - 2x^2 - 5x + 6$ is wasted effort, because $5$ doesn’t divide $6$.

Forgetting fractions when the leading coefficient isn’t 1. If no integer works for $2x^3 + x^2 + 5x - 3$, try $\pm\tfrac{1}{2}$ and $\pm\tfrac{3}{2}$ before deciding it can’t be factored.

Stopping too early. After dividing out one factor, check whether the quotient factors further. $(x - 1)(x^2 - x - 6)$ isn’t fully factored until you write $(x - 1)(x - 3)(x + 2)$.

Forcing a quadratic to factor. If the discriminant of the leftover quadratic is negative, it doesn’t factor over the real numbers. That’s a correct final answer, not a mistake.

Practice

1. (Warm-up) Let $P(x) = x^3 - x^2 - 4x - 6$. Is $x - 3$ a factor? Is $x + 2$ a factor?

Solution

$P(3) = 27 - 9 - 12 - 6 = 0$, so $x - 3$ is a factor.

$P(-2) = -8 - 4 + 8 - 6 = -10 \ne 0$, so $x + 2$ is not a factor.

2. (Warm-up) List the possible integer zeros of $x^3 + 2x^2 - 5x - 6$.

Solution

The factors of the constant term, $-6$: $\pm 1, \pm 2, \pm 3, \pm 6$.

3. (Core) Factor $x^3 + 2x^2 - 5x - 6$ fully.

Solution

Try $x = -1$: $P(-1) = -1 + 2 + 5 - 6 = 0$, so $x + 1$ is a factor.

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} -1 & 1 & 2 & -5 & -6 \\ & & -1 & -1 & 6 \\ \hline & 1 & 1 & -6 & \boxed{0} \end{array}$$

$x^2 + x - 6 = (x + 3)(x - 2)$, so

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

4. (Core) Factor $x^3 - 7x + 6$ fully.

Solution

$P(1) = 1 - 7 + 6 = 0$, so $x - 1$ is a factor. Use $0$ for the missing $x^2$ term:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & \boxed{0} \end{array}$$

$x^2 + x - 6 = (x + 3)(x - 2)$, so $x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3)$.

5. (Core) Find $k$ so that $x + 2$ is a factor of $x^3 + kx^2 - 2x + 8$. Then factor the polynomial as fully as possible.

Solution

$x + 2$ is a factor when $P(-2) = 0$:

$$-8 + 4k + 4 + 8 = 0 \quad\Rightarrow\quad 4k = -4 \quad\Rightarrow\quad k = -1$$

Now divide $x^3 - x^2 - 2x + 8$ by $x + 2$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} -2 & 1 & -1 & -2 & 8 \\ & & -2 & 6 & -8 \\ \hline & 1 & -3 & 4 & \boxed{0} \end{array}$$

$x^3 - x^2 - 2x + 8 = (x + 2)(x^2 - 3x + 4)$. The discriminant of $x^2 - 3x + 4$ is $9 - 16 = -7 \lt 0$, so it doesn’t factor further.

6. (Core) For $P(x) = 3x^3 + 4x^2 - 5x - 2$:

• (a) List the possible rational zeros.
• (b) Factor $P(x)$ fully.

Solution

(a) $\tfrac{b}{a}$ with $b$ a factor of $-2$ and $a$ a factor of $3$: $\pm 1, \pm 2, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}$.

(b) $P(1) = 3 + 4 - 5 - 2 = 0$, so $x - 1$ is a factor.

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 3 & 4 & -5 & -2 \\ & & 3 & 7 & 2 \\ \hline & 3 & 7 & 2 & \boxed{0} \end{array}$$

$3x^2 + 7x + 2 = (3x + 1)(x + 2)$, so $P(x) = (x - 1)(3x + 1)(x + 2)$.

7. (Core) Factor $x^4 - 2x^3 - 7x^2 + 8x + 12$ fully.

Solution

$P(-1) = 1 + 2 - 7 - 8 + 12 = 0$, so $x + 1$ is a factor.

$$\def\arraystretch{1.3} \begin{array}{r|rrrrr} -1 & 1 & -2 & -7 & 8 & 12 \\ & & -1 & 3 & 4 & -12 \\ \hline & 1 & -3 & -4 & 12 & \boxed{0} \end{array}$$

Factor the cubic $x^3 - 3x^2 - 4x + 12$ by grouping: $x^2(x - 3) - 4(x - 3) = (x - 3)(x^2 - 4) = (x - 3)(x - 2)(x + 2)$.

$$x^4 - 2x^3 - 7x^2 + 8x + 12 = (x + 1)(x - 3)(x - 2)(x + 2)$$

8. (Challenge) Find $a$ and $b$ so that both $x - 1$ and $x + 2$ are factors of $x^3 + ax^2 + bx - 6$. Then find the third factor.

Solution

$P(1) = 0$: $1 + a + b - 6 = 0$, so $a + b = 5$.

$P(-2) = 0$: $-8 + 4a - 2b - 6 = 0$, so $4a - 2b = 14$, or $2a - b = 7$.

Adding the two equations: $3a = 12$, so $a = 4$ and $b = 1$.

The polynomial is $x^3 + 4x^2 + x - 6$. Since $(x - 1)(x + 2) = x^2 + x - 2$, the third factor must be linear with leading coefficient $1$ and constant term $\tfrac{-6}{-2} = 3$. So it’s $x + 3$.

Check: $(x^2 + x - 2)(x + 3) = x^3 + 3x^2 + x^2 + 3x - 2x - 6 = x^3 + 4x^2 + x - 6$. ✓

9. (Challenge) Use the factor theorem to show that $x - a$ is a factor of $x^3 - a^3$ for any number $a$. Then find the other factor.

Solution

Let $P(x) = x^3 - a^3$. Then $P(a) = a^3 - a^3 = 0$, so $x - a$ is a factor.

Synthetic division with coefficients $1, 0, 0, -a^3$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} a & 1 & 0 & 0 & -a^3 \\ & & a & a^2 & a^3 \\ \hline & 1 & a & a^2 & \boxed{0} \end{array}$$

So $x^3 - a^3 = (x - a)(x^2 + ax + a^2)$. This is the difference of cubes pattern.