⚔️ Algebra Kingdom · Algebra

Factoring

Rewrite a polynomial as a product: pull out the greatest common factor, factor trinomials, and recognise the difference of squares.

In short

Factoring is expanding in reverse, so multiplying your factors back out always checks the answer.
Take out the greatest common factor before anything else — it makes every later step smaller.
$a 2$ - $b 2 = (a$ - b)(a $+ b)$ : two terms, subtracted, both perfect squares. A sum of squares does not factor.
To factor $x 2 +$ bx $+ c$ , find two numbers with product c and sum b; the signs of b and c tell you the signs of those numbers before you start searching.
"Factor completely" means no bracket can be broken down any further.
$a 3 + b 3 = (a + b)(a 2$ - ab $+ b 2)$ and $a 3$ - $b 3 = (a$ - $b)(a 2 +$ ab $+ b 2)$ ; SOAP gives the three signs — Same, Opposite, Always Positive.
Four terms with nothing shared by all of them is the signal to group: factor each pair, and the same bracket must come out of both.

Multiplying, run backwards

You already know how to expand: $(x + 2)(x + 5) = x 2 + 7x + 10$ . Factoring is the same journey in the other direction — starting from $x 2 + 7x + 10$ and recovering the two brackets it came from.

An area model shows why it works. Draw a rectangle whose sides are $(x + 2)$ and $(x + 5)$ . Cut it into four pieces and their areas are $x 2, 5x, 2x$ and 10. Add them up and you get $x 2 + 7x + 10$ . Factoring asks: given the four pieces, what were the two sides?

Factoring matters because a product tells you something a sum never can. If a product equals zero, one of its factors must be zero — and that is how quadratic equations get solved.

The check is always the same: multiply your factors back out. If you do not get what you started with, the factorisation is wrong.

Always take out the greatest common factor first

Before anything clever, look at every term and ask what they all share — both the numbers and the letters.

For $12x 3$ - $18x 2 + 30x$ :

the coefficients $12, 18$ and 30 all divide by 6
every term contains at least one x

So the greatest common factor is 6x, and $12x 3$ - $18x 2 + 30x = 6x(2x 2$ - $3x + 5)$ .

Two habits make this reliable. First, divide each term by the factor separately and write the results inside the bracket, so nothing is dropped. Second, check that the terms left inside share nothing further — if they do, you did not take out the *greatest* common factor.

Taking out the GCF first also makes everything that follows smaller and easier. $2x 2 + 10x + 12$ is much friendlier once it is written as $2(x 2 + 5x + 6)$ .

The difference of two squares

One pattern is worth memorising, because it appears everywhere:

$a 2$ - $b 2 = (a$ - b)(a $+ b)$

Expand the right-hand side and see why: the outer and inner products are $+ab$ and -ab, and they cancel, leaving $a 2$ - $b 2$ with no middle term at all.

To use it, check three things: there are exactly two terms, they are subtracted, and each one is a perfect square.

$x 2$ - $25 = (x$ - $5)(x + 5)$
$9x 2$ - $16 = (3x) 2$ - $42 = (3x$ - $4)(3x + 4)$
$4x 2$ - $81y 2 = (2x$ - $9y)(2x + 9y)$

A sum of squares such as $x 2 + 25$ does not factor with ordinary numbers, so do not force it. And note that $2x 2$ - 50 needs its common factor taken out first: $2(x 2$ - $25) = 2(x$ - $5)(x + 5)$ .

Trinomials, with and without a leading coefficient

Since $(x +$ m)(x $+ n) = x 2 + (m +$ n)x $+$ mn, factoring $x 2 +$ bx $+ c$ means finding two numbers whose product is c and whose sum is b.

The signs tell you most of the answer before you search:

c positive and b positive: both numbers positive. $x 2 + 7x + 10 = (x + 2)(x + 5)$
c positive and b negative: both numbers negative. $x 2$ - $7x + 10 = (x$ - 2)(x - 5)
c negative: one positive and one negative, and the bigger one carries the sign of $b. x 2$ - 3x - $10 = (x$ - $5)(x + 2)$

Then list the factor pairs of c and test their sums. For $c = 24$ the pairs are 1 and $24, 2$ and $12, 3$ and $8, 4$ and 6 — a short list, and one of them will have the sum you need.

Not every trinomial factors over the whole numbers. If no pair works, that is a real answer, not a mistake.

When the leading coefficient is not 1. For $ax 2 +$ bx $+ c$ with a other than 1, the a has to be split between the two brackets, and the ac-method organises the search.

Take $6x 2 + 7x$ - 3:

Multiply a by $c: 6 x (-3) = -18$ .
Find two numbers with product -18 and sum 7 (the middle coefficient): 9 and -2.
Split the middle term using them: $6x 2 + 9x$ - 2x - 3.
Factor each pair: $3x(2x + 3)$ - $1(2x + 3)$ .
The bracket $(2x + 3)$ is now common, so pull it out: $(2x + 3)(3x$ - 1).

Check by expanding: $6x 2$ - $2x + 9x$ - $3 = 6x 2 + 7x$ - 3.

"Factor completely" means keep going until nothing further can be pulled out. Take the common factor out first, then apply the pattern, then look once more at each bracket.

Two cubes, and four terms

Two more patterns finish the toolkit.

Sums and differences of cubes. Two terms, each one a perfect cube:

$a 3 + b 3 = (a + b)(a 2$ - ab $+ b 2) a 3$ - $b 3 = (a$ - $b)(a 2 +$ ab $+ b 2)$

Unlike a sum of squares, a sum of cubes really does factor. Take the cube root of each term, put those two roots into the binomial, and build the trinomial out of them: the first root squared, then the product of the roots, then the second root squared.

The signs are the only awkward part, and SOAP remembers them:

Same — the binomial takes the same sign as the original expression.
Opposite — the middle term of the trinomial takes the opposite sign.
Always Positive — the last term of the trinomial is always a plus.

So $x 3 + 27 = (x + 3)(x 2$ - $3x + 9)$ and $8x 3$ - $125 = (2x$ - $5)(4x 2 + 10x + 25)$ . That trinomial almost never breaks down further, so once it is written the work is done — but a common factor still comes out first: $2x 3$ - $54 = 2(x 3$ - $27) = 2(x$ - $3)(x 2 + 3x + 9)$ .

Four terms: group them. When a polynomial has four terms and nothing is shared by all four, split it into two pairs and factor each pair on its own:

$x 3 + 2x 2 + 3x + 6 = x 2 (x + 2) + 3(x + 2) = (x + 2)(x 2 + 3)$

The whole method rests on both pairs leaving the same bracket behind. If they do not, pair the terms up differently and try again. Watch the sign when a pair begins with a minus: in $x 3 + 2x 2$ - 3x - 6 the second pair gives $-3(x + 2)$ , not 3(x - 2).

Grouping also handles $ax 2 +$ bx $+ c$ . Split the middle term into the two numbers whose product is ac and whose sum is b, and three terms become four: $6x 2 + 7x$ - $3 = 6x 2 + 9x$ - 2x - $3 = 3x(2x + 3)$ - $1(2x + 3) = (2x + 3)(3x$ - 1).

Typing your answer. Write a product of brackets with nothing between them — $(x + 2)(x 2 + 3)$ — use $^$ for powers, and keep any factor that came out at the front, as in 2(x - $3)(x 2 + 3x + 9)$ .

Worked examples

Example 1

Factor completely: $12x 3$ - $18x 2 + 30x$

Look for a common factor. The coefficients $12, 18$ and 30 all divide by 6, and every term contains an x.
The greatest common factor is 6x.
Divide each term by $6x: 12x 3 / 6x = 2x 2, -18x 2 / 6x = -3x, 30x / 6x = 5$ .
So $12x 3$ - $18x 2 + 30x = 6x(2x 2$ - $3x + 5)$ .
The bracket $2x 2$ - $3x + 5$ has no common factor and does not factor further, so this is complete. Check by expanding: 6x times $2x 2$ is $12x 3$ , and so on.

Example 2

Factor completely: $x 2$ - 3x - 40

Two numbers are needed with product -40 and sum -3.
The product is negative, so one number is positive and the other negative; the sum is negative, so the larger one is the negative one.
Factor pairs of $40: 1$ and $40, 2$ and $20, 4$ and $10, 5$ and 8. The pair 5 and 8 has a difference of 3.
Take -8 and $+5: (-8) x 5 = -40$ and $-8 + 5 = -3$ . Correct.
$x 2$ - 3x - $40 = (x$ - $8)(x + 5)$ . Check: $x 2 + 5x$ - 8x - $40 = x 2$ - 3x - 40.

Example 3

Factor completely: $2x 2$ - 50

Both terms divide by 2, so take that out first: $2(x 2$ - 25).
The bracket has two terms, subtracted, and both are perfect squares — a difference of squares.
$x 2$ - $25 = (x) 2$ - $(5) 2 = (x$ - $5)(x + 5)$ .
So $2x 2$ - $50 = 2(x$ - $5)(x + 5)$ .
Check: (x - $5)(x + 5) = x 2$ - 25, and doubling gives $2x 2$ - 50.

Example 4

Factor completely: $8x 3$ - 125

Two terms, and neither one is a perfect square — so check for cubes. $8x 3 = (2x) 3$ and $125 = 5 3$ .
This is a difference of cubes: $a 3$ - $b 3 = (a$ - $b)(a 2 +$ ab $+ b 2)$ , with $a = 2x$ and $b = 5$ .
The binomial is (2x - 5), taking the Same sign as the original.
The trinomial is $(2x) 2 + (2x)(5) + 5 2 = 4x 2 + 10x + 25$ — Opposite sign in the middle, Always Positive at the end.
So $8x 3$ - $125 = (2x$ - $5)(4x 2 + 10x + 25)$ . Expanding gives $8x 3 + 20x 2 + 50x$ - $20x 2$ - 50x - 125, and everything in the middle cancels.

Example 5

Factor completely: $2x 3$ - $6x 2 + 5x$ - 15

Four terms, and nothing is shared by all four, so group them in pairs: $(2x 3$ - $6x 2) + (5x$ - 15).
The first pair shares $2x 2 : 2x 3$ - $6x 2 = 2x 2 (x$ - 3).
The second pair shares $5: 5x$ - $15 = 5(x$ - 3).
Both pairs left the bracket (x - 3) behind, so take it out of both: (x - $3)(2x 2 + 5)$ .
Check by expanding: $2x 3$ - $6x 2 + 5x$ - 15, which is what we started with.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Factor completely. $15x + 25$ Write your answer as a product of factors.

Answer: $5(3x + 5)$

Coefficients: $15, 25$ — their greatest common factor is 5.
The terms do not all contain an x, so only the number 5 comes out.
Divide every term by 5; what is left inside the bracket is $3x + 5$ .
Answer: $5(3x + 5)$
Check: multiplying back gives $15x + 25$ .

Problem 2

Difficulty 3 of 5

Factor completely. $25x 2$ - 9 Write your answer as a product of factors.

Answer: (5x - $3)(5x + 3)$

$25x 2$ - $9 = (5x) 2$ - $(3) 2$
A difference of two squares factors as (a - b)(a $+ b)$ .
Answer: (5x - $3)(5x + 3)$
Check: the two middle terms are -15x and $+15x$ , so they cancel and leave $25x 2$ - 9.

Problem 3

Difficulty 4 of 5

Factor completely. $x 2$ - $27x + 180$ Write your answer as a product of factors.

Answer: (x - 12)(x - 15)

Look for m and n with $m x n = 180$ and $m + n = -27$ .
$-12 x -15 = 180$ and $-12 + -15 = -27$ , so $m = -12$ and $n = -15$ .
Answer: (x - 12)(x - 15)
Check: (x - 12)(x - 15) expands to $x 2$ - $27x + 180$ .

Common mistakes

Stopping too early — taking out the common factor but leaving a bracket like $x 2$ - 9 that still factors.
Taking out only the number and forgetting the shared x, writing $3(2x 2 + 3x)$ instead of $3x(2x + 3)$ .
Losing the common factor from the answer: writing $(x + 2)(x + 3)$ when the polynomial was $2x 2 + 10x + 12$ .
Getting the signs the wrong way round, so that (x - 2)(x - 5) is offered for $x 2 + 7x + 10$ .
Trying to factor a sum of squares such as $x 2 + 16$ as if it were a difference.
Giving the trinomial of a cube factorisation the wrong middle sign, so $x 3 + 27$ comes out as $(x + 3)(x 2 + 3x + 9)$ . SOAP says Opposite.
Writing the second factor of a cube as a perfect square — $(x + 3)(x 2 + 6x + 9)$ — when the middle term is 3x, not 6x.
Losing the sign of the second group, so $x 3 + 2x 2$ - 3x - 6 is offered as $(x + 2)(x 2 + 3)$ instead of $(x + 2)(x 2$ - 3).

What you should be able to do

Factor out the greatest common factor of a polynomial.
Factor a trinomial of the form $x 2 +$ bx $+ c$ .
Recognise and factor a difference of two squares.
Multiply the factors back to check the result.

Where this fits in the curriculum

Common Core

HSA-SSE.A.2
High school — Use the structure of an expression to rewrite it, such as seeing a difference of two squares.
HSA-SSE.B.3.A
High school — Factor a quadratic expression to reveal the zeros of the function it defines.
HSA-APR.A.1
High school — Understand that polynomials are closed under addition, subtraction and multiplication (multiplying the factors back).

SAT

Passport to Advanced Math
Factoring quadratics and using the structure of an expression.