How to factor: GCF, trinomials and difference of squares

Q: Can every polynomial be factored?

No. Some are prime over the real numbers, like x^2 + 9 or x^2 + x + 1. Others factor only with irrational numbers, like x^2 - 2 = (x + \sqrt{2})(x - \sqrt{2}). If the quick check b^2 - 4ac is not a perfect square, whole-number factoring will not work.

The key idea

Multiplying turns a product into a sum. Factoring runs that backward:

(x + 4)(x + 5) \;\xrightarrow{\text{multiply}}\; x^2 + 9x + 20 \;\xrightarrow{\text{factor}}\; (x + 4)(x + 5)

The 4 and 5 multiply to 20 (the last number) and add to 9 (the middle number). That pattern is the whole trick for $x^2 + bx + c$ . Use this checklist every time:

Take out the greatest common factor (GCF), if there is one.
Count the terms. Two terms: look for a difference of squares. Three terms: use the product and sum. Four terms: try grouping.
Check whether any factor can be factored again.
Multiply back to check.

Signs for x^2 + bx + c
If c is	and b is	the two numbers are	Example
positive	positive	both positive	$x^2 + 9x + 20 = (x + 4)(x + 5)$
positive	negative	both negative	$x^2 - 9x + 20 = (x - 4)(x - 5)$
negative	either	opposite signs; the larger one has b's sign	$x^2 - x - 20 = (x - 5)(x + 4)$

Worked examples

Example 1: greatest common factor

Problem Factor $8x^3y - 12x^2y^2$ .

The GCF of 8 and 12 is 4. Both terms have at least $x^2$ and at least $y$ . So the GCF is $4x^2y$ . Divide each term by it.
$8x^3y - 12x^2y^2 = 4x^2y(2x - 3y)$
Check: $4x^2y \cdot 2x = 8x^3y$ and $4x^2y \cdot 3y = 12x^2y^2$ .

Answer $4x^2y(2x - 3y)$

Example 2: a trinomial with a = 1

Problem Factor $x^2 - 2x - 35$ .

Find two numbers that multiply to $-35$ and add to $-2$ . Since $-35$ is negative, the signs are opposite. The pairs are 1 and 35, or 5 and 7. The pair $-7$ and $5$ adds to $-2$ .
Write the factors.
$x^2 - 2x - 35 = (x - 7)(x + 5)$
Check with FOIL: $x^2 + 5x - 7x - 35 = x^2 - 2x - 35$ .

Answer $(x - 7)(x + 5)$

Example 3: a trinomial with a not 1 (the ac method)

Problem Factor $6x^2 + 7x - 3$ .

Multiply $a \cdot c = 6 \cdot (-3) = -18$ . Find two numbers that multiply to $-18$ and add to 7: they are 9 and $-2$ . Use them to split the middle term.
$6x^2 + 7x - 3 = 6x^2 + 9x - 2x - 3$
Group the first two and the last two terms, and take the GCF of each group. Factor $-1$ out of the second group so both groups share $(2x + 3)$ .
$6x^2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3)$
Factor out the shared $(2x + 3)$ .
$3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3)$

Answer $(3x - 1)(2x + 3)$

Example 4 (test-hard): factor completely

Problem Factor $2x^4 - 32$ completely.

GCF first.
$2x^4 - 32 = 2(x^4 - 16)$
$x^4 - 16$ is $(x^2)^2 - 4^2$ , a difference of squares.
$2(x^4 - 16) = 2(x^2 + 4)(x^2 - 4)$
$x^2 - 4$ is another difference of squares. $x^2 + 4$ is a sum of squares, which does not factor with real numbers.
$2(x^2 + 4)(x^2 - 4) = 2(x^2 + 4)(x + 2)(x - 2)$

Answer $2(x^2 + 4)(x + 2)(x - 2)$

Common mistakes

Skipping the GCF. $2x^2 + 10x + 12$ is much easier as $2(x^2 + 5x + 6) = 2(x + 2)(x + 3)$ . Fix: always look for a GCF first.
Right numbers, wrong signs. $(x + 7)(x - 5)$ gives $+2x$ , not $-2x$ . Fix: the larger number gets the sign of b.
Trying to factor a sum of squares. $x^2 + 9$ does not factor over the real numbers. Only a difference of squares does.
Stopping too early. $(x^2 + 4)(x^2 - 4)$ is not finished, because $x^2 - 4$ still factors.
Sign slip in grouping. Taking $-1$ out of $-2x - 3$ gives $-1(2x + 3)$ , with a plus inside.

Quick methods

Practice

5 practice questions

Factor $x^2 + 9x + 20$ .
1. $(x + 4)(x + 5)$
2. $(x + 2)(x + 10)$
3. $(x - 4)(x - 5)$
4. $(x + 1)(x + 20)$
Show answer
Answer: $(x + 4)(x + 5)$
4 and 5 multiply to 20 and add to 9. The pair 2 and 10 multiplies to 20 but adds to 12. $(x - 4)(x - 5)$ gives $-9x$ .
Factor $x^2 - 49$ .
1. $(x - 7)^2$
2. $(x + 7)(x - 7)$
3. $(x + 7)^2$
4. $(x - 49)(x + 1)$
Show answer
Answer: $(x + 7)(x - 7)$
It is a difference of squares, $x^2 - 7^2$ . $(x - 7)^2$ multiplies out to $x^2 - 14x + 49$ , which has a middle term and a plus 49.
Factor $3x^2 - 10x - 8$ .
1. $(3x + 2)(x - 4)$
2. $(3x - 2)(x + 4)$
3. $(3x - 4)(x + 2)$
4. $(3x + 4)(x - 2)$
Show answer
Answer: $(3x + 2)(x - 4)$
$ac = -24$ , and $-12$ and 2 multiply to $-24$ and add to $-10$ . Split: $3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4)$ . The other choices give middle terms of $+10x$ , $+2x$ and $-2x$ .
Which shows $5x^3 - 20x$ factored completely?
1. $5x(x^2 - 4)$
2. $5x(x + 2)(x - 2)$
3. $5(x^3 - 4x)$
4. $5x(x - 2)^2$
Show answer
Answer: $5x(x + 2)(x - 2)$
Take out the GCF $5x$ to get $5x(x^2 - 4)$ . That is correct but not complete: $x^2 - 4$ is a difference of squares. $5(x^3 - 4x)$ misses part of the GCF, and $5x(x - 2)^2$ is a different polynomial.
Which is a factor of $2x^2 + 5x - 12$ ?
1. $2x + 3$
2. $x - 4$
3. $x + 4$
4. $2x + 4$
Show answer
Answer: $x + 4$
$ac = -24$ , and 8 and $-3$ add to 5. Split and group: $2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)$ . So $x + 4$ is a factor. The signs in $2x + 3$ and $x - 4$ are flipped.

Frequently asked questions

What is the first step in factoring?

Look for a greatest common factor and take it out. It makes every later step smaller and easier. Then count the terms: two terms suggest a difference of squares, three suggest a trinomial, and four suggest grouping.

How do I factor a trinomial when a is not 1?

Use the ac method. Multiply a and c, then find two numbers that multiply to $ac$ and add to b. Split the middle term into those two pieces, group the four terms into pairs, factor each pair, and pull out the shared binomial.

Can every polynomial be factored?

No. Some are prime over the real numbers, like $x^2 + 9$ or $x^2 + x + 1$ . Others factor only with irrational numbers, like $x^2 - 2 = (x + \sqrt{2})(x - \sqrt{2})$ . If the quick check $b^2 - 4ac$ is not a perfect square, whole-number factoring will not work.

How do I check my factoring?

Multiply the factors back together. If you get the original polynomial, your answer is right. Then make sure no factor can be factored again, since "factor completely" means every piece is as small as it can be.

How to factor polynomials

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

Try asking Ducky

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

Related

Try asking Ducky