Factor algebraic expressions

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 30 minutes
Domains: Advanced Math
Techniques: FactoringGreatest-common-factorDifference-of-squaresQuadratic-trinomialsFactoring-by-grouping

What you’ll learn

Tell when a question wants an expression written as a product.
Use graph overlap in Desmos when the answer choices are complete products in one variable and matching them is quicker than factoring.
Take out the greatest common factor before you look for any other pattern.
Factor differences of squares, quadratic trinomials and four-term expressions by grouping.
Treat a repeated chunk, like $(x-1)$ , as one temporary quantity.
Expand to check a factorization you did by hand, when you need to.

Prerequisites

Why this matters on the SAT

Find the matching product without expanding every choice

Some SAT questions show a long polynomial and ask which answer choice is the same expression written as a product. You could factor the polynomial yourself, or expand all four choices and compare. But when no short pattern jumps out, Desmos can do the comparing for you: the right choice's graph lands exactly on top of the original's.

SAT example

Which expression is equivalent to

2x^4+7x^3-3x^2-2x-24?

A
$(2x-3)(x+4)(x^2+x+2)$
B
$(2x+3)(x-4)(x^2+x+2)$
C
$(2x-3)(x-4)(x^2-x+2)$
D
$(2x+3)(x+4)(x^2-x+2)$

Fast Desmos solution

Type the original on line 1 and leave it there. Type choice A on line 2. If it doesn't match, edit line 2 into the next choice instead of adding new lines.

With choice A on line 2, the graph changes color but doesn't move. The two curves overlap everywhere in the window, so the answer is A.

You don't need to expand choice A to double-check. The choices are already exact products, and a clear match picks one out. If two choices ever look the same on screen, zoom in or compare their algebra rather than guessing from the picture.

Calculator loads as you approach

Toggle line 2 off and on. The curve doesn’t move, because choice A overlaps the original everywhere in the window.

Decide whether to graph or factor by hand

Factoring rewrites a sum or difference as a product. It's distribution run backward:

ab+ac=a(b+c).

Read left to right, you pull the shared $a$ out of both terms.

Before any algebra, look at what the question wants. Some questions give you four products and ask which one matches. Others ask you to build the product yourself. Matching can often go to Desmos. Building is a job for your pencil.

Pick whichever way has less setup and fewer places to slip.

Start with graph overlap when…

Each choice is a whole product, like $(2x-3)(x+4)(x^2+x+2)$ , so you can type it into Desmos as it is.
Everything uses just one variable, like $x$ , and the coefficients are plain numbers, not letters like $k$ .
You look for a shared factor or another quick pattern and don’t see one.

Factor by hand when…

There are no complete products to choose from, so you have to write the product yourself.
A quick pattern jumps out, like the $2x$ that both terms of $18x^3-50x$ share.
The answer keeps more than one variable, like $6x^2y(2x-3y)$ , or a letter like $k$ that can stand for any number.

By hand, go one layer at a time: pull out one factor, then look again at what’s left. Expand at the end only when you need to check.

Two other questions look similar but want something different. If a question only asks whether one piece, like $x-2$ , is a factor, you don't need the whole factorization. A factor or root test is quicker: $x-2$ is a factor exactly when the expression equals $0$ at $x=2$ . And if an equation is set equal to zero and asks for values of the variable, turn to Solve quadratic equations by factoring.

For more practice on the Desmos side, try Equivalent expressions by graph overlap.

Start with the greatest common factor

The greatest common factor, or GCF, is the biggest factor that every term shares, numbers and variables together. You find it in two parts:

For the numbers, take the largest number that divides every coefficient.
For each variable, take the smallest power of it that appears in every term.

Take $12x^3y-18x^2y^2$ . The largest number that divides both $12$ and $18$ is $6$ . The smallest power of $x$ is $x^2$ , and the smallest power of $y$ is $y$ . So the GCF is $6x^2y$ . Divide each term by it:

\begin{aligned} 12x^3y-18x^2y^2 &=6x^2y\left(\frac{12x^3y}{6x^2y}-\frac{18x^2y^2}{6x^2y}\right)\\[1.4em] &=6x^2y(2x-3y). \end{aligned}

Where did the $x$ in $2x$ come from? Taking $x^2$ out of $x^3$ leaves $x^{3-2}=x$ . That's the exponent rule from Use exponent rules and common bases.

Check your understanding:

What’s the greatest common factor of $20a^4b^2-30a^3b^5$ , and what’s left after you factor it out?

Common mistake:

A common factor has to come out of every term. Writing $6x^2+4x+5$ as $2x(3x+2+5)$ changes the expression, because the $5$ has no factor of $2x$ to give up. To catch this, distribute your factor back through every term and compare the result with the original.

Use the difference of squares

A difference of squares is exactly two terms, one subtracted from the other, where each term is a square:

A^2-B^2=(A-B)(A+B).

Why does that work? Expand the right side, using the multiplication from Distribute, combine, and rewrite expressions:

\begin{aligned} (A-B)(A+B) &=A^2+AB-AB-B^2\\[1.4em] &=A^2-B^2. \end{aligned}

The middle terms, $+AB$ and $-AB$ , cancel, so only the two squares are left.

Sometimes the pattern stays hidden until you take out the GCF. Neither term of $18x^3-50x$ is a square, but both share $2x$ . Take that out first:

\begin{aligned} 18x^3-50x &=2x(9x^2-25)\\[1.4em] &=2x\big((3x)^2-5^2\big)\\[1.4em] &=2x(3x-5)(3x+5). \end{aligned}

Watch the sign, though. A sum of squares, like $A^2+B^2$ , doesn't factor with this pattern over the real numbers.

Check your understanding:

Why is $25p^4-4q^2$ a difference of squares, and how does it factor?

Factor a quadratic trinomial

A quadratic trinomial has three terms: an $x^2$ term, an $x$ term and a number. In general, it looks like

ax^2+bx+c,

where $a\ne0$ . To see how to take one apart, first watch one get built. Multiply two binomials:

(3x+2)(2x-1)=6x^2-3x+4x-2=6x^2+x-2.

Here $a=6$ , $b=1$ and $c=-2$ . The first terms, $3x$ and $2x$ , gave $6x^2$ . The last terms, $2$ and $-1$ , gave $-2$ . The middle term came from the two cross products, $3x$ times $-1$ and $2$ times $2x$ , which gave $-3x$ and $4x$ . Look at the numbers in those two pieces, $-3$ and $4$ :

They add to $1$ , which is $b$ .
They multiply to $-12$ , which is $6$ times $-2$ , or $ac$ .

That second fact isn't luck. Between them, $-3=3\cdot(-1)$ and $4=2\cdot2$ use all four numbers from the binomials: $3$ , $-1$ , $2$ and $2$ . And $ac=(3\cdot2)\big(2\cdot(-1)\big)$ multiplies the same four numbers. So factoring runs this backward: find two numbers that multiply to $ac$ and add to $b$ .

When $a=1$ , $ac$ is the same as $c$ , so you need two numbers that multiply to $c$ and add to $b$ . For $x^2-2x-15$ , they're $-5$ and $3$ :

\begin{aligned} x^2-2x-15 &=(x-5)(x+3), \end{aligned}

because $(-5)(3)=-15$ and $-5+3=-2$ .

If the same number shows up twice, you can write the product as a square:

x^2+6x+9=(x+3)(x+3)=(x+3)^2.

When $a\ne1$ , there's one more move. You use the two numbers to split the middle term into two pieces, then group. Try it on the trinomial from before,

6x^2+x-2.

You need two numbers whose product is

ac=(6)(-2)=-12

and whose sum is $b=1$ . That's $4$ and $-3$ , so split $x$ into $4x-3x$ :

\begin{aligned} 6x^2+x-2 &=6x^2+4x-3x-2\\[1.4em] &=2x(3x+2)-1(3x+2)\\[1.4em] &=(3x+2)(2x-1). \end{aligned}

Both groups share $3x+2$ , so it comes out as one factor. And you're back to the product you started with, which is your check: $(3x+2)(2x-1)$ expands to $6x^2+x-2$ .

Try it yourself:

Before you open the check, find the pair for $3x^2-14x-5$ . It has to multiply to $-15$ and add to $-14$ .

Check your understanding:

Which pair works, and how does it lead to the factored form of $3x^2-14x-5$ ?

Group terms and keep factoring

With four terms, try grouping them in pairs. You want each pair to leave the same expression after you factor it:

\begin{aligned} 4x^3+12x^2-9x-27 &=(4x^3+12x^2)+(-9x-27)\\[1.4em] &=4x^2(x+3)-9(x+3)\\[1.4em] &=(x+3)(4x^2-9). \end{aligned}

Both pairs left $x+3$ , so it comes out as a factor.

Don't stop yet. Look again at what's left: $4x^2-9$ is a difference of squares.

4x^2-9=(2x-3)(2x+3).

So the complete factorization is

\boxed{(x+3)(2x-3)(2x+3)}.

Common mistake:

It’s easy to stop at $(x+3)(4x^2-9)$ , because it already looks like a product. But $4x^2-9$ still factors. After every step, check each factor that still has a variable in it. Then expand your final product to make sure no sign or term changed along the way.

Treat a repeated expression as one quantity

Sometimes the same chunk shows up more than once, like $x-1$ in $2(x-1)^2+5(x-1)-3$ . Look at the shape: the chunk squared, the chunk itself, then a number. That's the shape of a trinomial, with $x-1$ where $x$ usually goes. Give the chunk a temporary name, and the trinomial is easier to see.

Let

u=x-1.

Then

2(x-1)^2+5(x-1)-3=2u^2+5u-3.

Now it's an ordinary trinomial in $u$ . You need two numbers that multiply to $ac=(2)(-3)=-6$ and add to $5$ . That's $6$ and $-1$ , so split $5u$ into $6u-u$ and group:

\begin{aligned} 2u^2+5u-3 &=2u^2+6u-u-3\\[1.4em] &=2u(u+3)-1(u+3)\\[1.4em] &=(2u-1)(u+3). \end{aligned}

Last, put $x-1$ back in for $u$ , in both factors:

\begin{aligned} (2(x-1)-1)((x-1)+3) &=(2x-3)(x+2). \end{aligned}

The letter $u$ only keeps the work tidy. The expression itself never changes.

Example: Factor in layers

Worked example

Which expression is equivalent to

6x^3-15x^2-9x?

A
$3x(2x+1)(x-3)$
B
$3x(2x-1)(x+3)$
C
$3(2x+1)(x-3)$
D
$3x(2x+3)(x-1)$

Step 1

Take out the GCF

The coefficients $6$ , $15$ and $9$ all share $3$ , and every term has an $x$ . So pull out $3x$ :

6x^3-15x^2-9x=3x(2x^2-5x-3).

Step 2

Look at what’s left

The trinomial $2x^2-5x-3$ might factor too. Its $ac$ is

(2)(-3)=-6,

so you need two numbers that multiply to $-6$ and add to $-5$ . That's $-6$ and $1$ .

Step 3

Split and group

Use the pair to split $-5x$ into $-6x+x$ , then group:

\begin{aligned} 2x^2-5x-3 &=2x^2-6x+x-3\\[1.4em] &=2x(x-3)+1(x-3)\\[1.4em] &=(2x+1)(x-3). \end{aligned}

Step 4

Bring back the GCF and check

Don't forget the $3x$ from the first step:

6x^3-15x^2-9x=3x(2x+1)(x-3).

Expand to check:

\begin{aligned} 3x(2x+1)(x-3) &=3x(2x^2-5x-3)\\[1.4em] &=6x^3-15x^2-9x. \end{aligned}

The answer is A.

Common mistake:

Choice C has the right trinomial factors but loses the $x$ from the GCF. Keep every factor you pull out on each line you write. A quick look at the highest power catches this too: choice C expands to an expression whose highest power is $x^2$ , but the original has $x^3$ .

Practice problems

Before you start each one, decide: graph the choices, or factor by hand?

Take out the GCF, then spot the squares

Practice problem

Which expression is equivalent to

12t^3-75t?

Calculator loads as you approach

Use the calculator only if it helps you check a choice. The common factor and the difference of squares are quicker by hand.

Match a long product by graph overlap

Practice problem

Which expression is equivalent to

6x^3-19x^2+11x+6?

Calculator loads as you approach

Keep the original on line 1, and try one complete choice at a time on line 2.

Factor a repeated quadratic chunk

Practice problem

The expression

6(x^2+x-1)^2-(x^2+x-1)-2

can be rewritten as

(3x^2+3x+a)(2x^2+2x+b),

where $a$ and $b$ are integers. What is the value of $a+b$ ?

Your answer

Calculator loads as you approach

Name the repeated chunk first. The calculator is here if you want to check your expansion at the end.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Factoring runs distribution backward: it turns a sum or difference into a product.
Long product choices in one variable, and no pattern in sight? Graph them and look for the one that lands on top. Otherwise, factor by hand.
Take out the greatest common factor first. After every step, look again at what's left, until no factor fits one of these patterns.
A difference of squares needs two squares with a minus between them: $A^2-B^2=(A-B)(A+B)$ .
Use the trinomial method only when the powers form a quadratic in one quantity: its square, the quantity itself, then a number. Find two numbers that multiply to $ac$ and add to $b$ . If that quantity is a chunk, like $x-1$ in $2(x-1)^2+5(x-1)-3$ , call the chunk $u$ first, then put it back for every $u$ .
With four terms, group them in pairs that leave the same factor.
Expand to check a hand factorization, or to settle a graph that's too close to call.

Related lessons

When an equation is set equal to zero, go to Solve quadratic equations by factoring. When a given identity or factor pins down unknown constants, go to Use polynomial identities, factors, and unknown coefficients.

Next lesson

Rewrite rational expressions and preserve restrictions

Use complete factorization to cancel factors while keeping every denominator restriction.

Start next lesson

Practice

Practice this lesson

491 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice