Factoring Polynomials

You’ve met lots of factoring methods over the years. For cubics and quartics, the hard part is often choosing which method to use. This page puts all the tools in one place with a checklist for picking the right one. Factoring is how you’ll solve polynomial equations and inequalities next, so it pays to be quick and reliable.

Key ideas

The toolkit

Method	Pattern	Example
Common factor	$ab + ac = a(b + c)$	$3x^3 - 12x = 3x(x^2 - 4)$
Difference of squares	$a^2 - b^2 = (a - b)(a + b)$	$x^2 - 25 = (x - 5)(x + 5)$
Trinomial	$x^2 + bx + c = (x + r)(x + s)$ with $rs = c$, $r + s = b$	$x^2 - x - 6 = (x - 3)(x + 2)$
Grouping	factor pairs of terms, then the common binomial	$x^3 + 2x^2 + 3x + 6 = (x + 2)(x^2 + 3)$
Difference of cubes	$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$	$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
Sum of cubes	$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$	$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$
Factor theorem	$P(a) = 0$ means $x - a$ is a factor	see the factor theorem

Sum and difference of cubes

These two patterns are new for most students. A way to remember the signs is SOAP: the signs in the answer go Same, Opposite, Always Positive.

$$a^3 - b^3 = (a \;\underset{\text{same}}{-}\; b)(a^2 \;\underset{\text{opposite}}{+}\; ab \;\underset{\text{always +}}{+}\; b^2)$$

To use them, write each term as a cube: $8x^3 = (2x)^3$, $27 = 3^3$, $64 = 4^3$. The quadratic factor $a^2 \pm ab + b^2$ never factors further over the real numbers.

Note there is no sum of squares pattern: $x^2 + 9$ doesn’t factor over the real numbers.

Quadratic-type quartics

A quartic with only $x^4$, $x^2$ and constant terms, like $x^4 - 13x^2 + 36$, is a quadratic in disguise. Think of $x^2$ as a single unknown (you can write $u = x^2$ if it helps):

$$x^4 - 13x^2 + 36 = (x^2)^2 - 13(x^2) + 36 = (x^2 - 4)(x^2 - 9)$$

Each bracket is a difference of squares, so the full factorization is $(x - 2)(x + 2)(x - 3)(x + 3)$.

A checklist for choosing a strategy

1. Common factor first. Always take out the greatest common factor (including $-1$ if the leading coefficient is negative and that helps).
2. Count the terms.
   • Two terms: difference of squares, or sum or difference of cubes.
   • Three terms: trinomial factoring, including quadratic-type quartics.
   • Four terms: try grouping.
3. Still stuck? Use the factor theorem: test the possible zeros, then divide.
4. Check every factor to see whether it factors again. Stop when no factor can be broken down further.
5. Check your answer by expanding, or by substituting a value like $x = 1$ into both forms.

Worked examples

Example 1: Common factor, then a trinomial

Factor $2x^3 - 8x^2 - 24x$.

Solution. Every term has a factor of $2x$:

$$2x^3 - 8x^2 - 24x = 2x(x^2 - 4x - 12)$$

Now find two numbers that multiply to $-12$ and add to $-4$: $-6$ and $2$.

$$2x^3 - 8x^2 - 24x = 2x(x - 6)(x + 2)$$

Check with $x = 1$: the left side is $2 - 8 - 24 = -30$, and the right side is $2(1)(-5)(3) = -30$. ✓

Example 2: Grouping

Factor $x^3 + 3x^2 - 4x - 12$.

Solution. Four terms, so try grouping. Take a common factor out of each pair:

$$\begin{aligned} x^3 + 3x^2 - 4x - 12 &= x^2(x + 3) - 4(x + 3) \\ &= (x + 3)(x^2 - 4) && \text{common factor } x + 3 \\ &= (x + 3)(x - 2)(x + 2) && \text{difference of squares} \end{aligned}$$

Notice the second pair: factoring out $-4$ (not $4$) makes the bracket $x + 3$ match the first one.

Example 3: A quadratic-type quartic

Factor $x^4 - 10x^2 + 9$.

Solution. Treat $x^2$ as the unknown. Find two numbers that multiply to $9$ and add to $-10$: $-1$ and $-9$.

$$\begin{aligned} x^4 - 10x^2 + 9 &= (x^2 - 1)(x^2 - 9) \\ &= (x - 1)(x + 1)(x - 3)(x + 3) \end{aligned}$$

Check with $x = 2$: the left side is $16 - 40 + 9 = -15$, and the right side is $(1)(3)(-1)(5) = -15$. ✓

Example 4: Common factor, then a sum of cubes

Factor $2x^4 + 54x$.

Solution. Take out the common factor $2x$:

$$2x^4 + 54x = 2x(x^3 + 27)$$

Now $x^3 + 27 = x^3 + 3^3$ is a sum of cubes with $a = x$ and $b = 3$:

$$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$$

So

$$2x^4 + 54x = 2x(x + 3)(x^2 - 3x + 9)$$

The quadratic $x^2 - 3x + 9$ has discriminant $9 - 36 = -27 \lt 0$, so it doesn’t factor further.

Common mistakes

Skipping the common factor. If you start factoring $2x^3 - 8x^2 - 24x$ without taking out $2x$, everything is harder. Always look for a common factor first.

Getting the cube signs wrong. $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$, not $(x - 2)(x^2 - 2x + 4)$ or $(x - 2)(x + 2)^2$. Use SOAP: same, opposite, always positive.

Trying to factor a sum of squares. $x^2 + 9$ and $x^2 + 4$ don’t factor over the real numbers. $x^4 - 81 = (x^2 - 9)(x^2 + 9)$ factors once more as $(x - 3)(x + 3)(x^2 + 9)$, but the $x^2 + 9$ stays.

Sign errors in grouping. In $x^3 + 3x^2 - 4x - 12$, the second pair is $-4x - 12 = -4(x + 3)$. Writing $-4(x - 3)$ breaks the pattern. Expand each pair back in your head to check.

Not factoring fully. $(x^2 - 1)(x^2 - 9)$ is a correct step, but it isn’t finished. Each bracket is a difference of squares. Keep going until no factor can be broken down.

Practice

1. (Warm-up) Factor $3x^3 - 12x$.

Solution

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

2. (Warm-up) Factor $x^4 - 81$.

Solution

$$x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$$

$x^2 + 9$ is a sum of squares, so it doesn’t factor further.

3. (Warm-up) Factor $x^3 - 64$.

Solution

$64 = 4^3$, so use the difference of cubes with $a = x$ and $b = 4$:

$$x^3 - 64 = (x - 4)(x^2 + 4x + 16)$$

4. (Core) Factor $x^3 - 2x^2 - 9x + 18$.

Solution

$$\begin{aligned} x^3 - 2x^2 - 9x + 18 &= x^2(x - 2) - 9(x - 2) \\ &= (x - 2)(x^2 - 9) \\ &= (x - 2)(x - 3)(x + 3) \end{aligned}$$

5. (Core) Factor $x^4 - 17x^2 + 16$.

Solution

Two numbers that multiply to $16$ and add to $-17$: $-1$ and $-16$.

$$\begin{aligned} x^4 - 17x^2 + 16 &= (x^2 - 1)(x^2 - 16) \\ &= (x - 1)(x + 1)(x - 4)(x + 4) \end{aligned}$$

6. (Core) Factor $5x^3 + 40$.

Solution

$$5x^3 + 40 = 5(x^3 + 8) = 5(x + 2)(x^2 - 2x + 4)$$

7. (Core) Factor $2x^3 + 3x^2 - 11x - 6$. (Grouping doesn’t work here.)

Solution

Possible rational zeros: $\pm 1, \pm 2, \pm 3, \pm 6, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$. Try $x = 2$: $16 + 12 - 22 - 6 = 0$, so $x - 2$ is a factor.

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & & 4 & 14 & 6 \\ \hline & 2 & 7 & 3 & \boxed{0} \end{array}$$

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

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

8. (Challenge) Factor $x^4 + x^3 - 8x - 8$.

Solution

Group in pairs, then use the difference of cubes:

$$\begin{aligned} x^4 + x^3 - 8x - 8 &= x^3(x + 1) - 8(x + 1) \\ &= (x + 1)(x^3 - 8) \\ &= (x + 1)(x - 2)(x^2 + 2x + 4) \end{aligned}$$

9. (Challenge) Factor $4x^4 - 17x^2 + 4$.

Solution

Treat $x^2$ as the unknown: $4u^2 - 17u + 4$ with $u = x^2$. Two numbers that multiply to $4 \times 4 = 16$ and add to $-17$ are $-1$ and $-16$:

$$\begin{aligned} 4x^4 - 17x^2 + 4 &= 4x^4 - x^2 - 16x^2 + 4 \\ &= x^2(4x^2 - 1) - 4(4x^2 - 1) \\ &= (4x^2 - 1)(x^2 - 4) \\ &= (2x - 1)(2x + 1)(x - 2)(x + 2) \end{aligned}$$