Factoring Polynomials: GCF, Trinomials, and Difference of Squares

Recommended resources

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Good to see you again. Last lesson focused on expanding polynomials: distributing, combining like terms, and using FOIL for binomials. Factoring reverses that process. Instead of multiplying factors to create a polynomial, you start with a polynomial and rewrite it as a product of simpler factors.

In this lesson, you will learn a reliable order for factoring with the three most important high-school patterns: greatest common factor, factorable trinomials, and difference of squares. This matters immediately because the next lesson will use these factored forms to solve quadratic equations.

Factoring begins with structure

For example, multiplication tells us that

3 x (x - 4) = 3 x^{2} - 12 x .

Factoring reverses it:

3 x^{2} - 12 x = 3 x (x - 4) .

Both expressions have exactly the same value for every value of $x$ , but the factored form reveals useful structure. The expressions $3 x$ and $(x - 4)$ are called factors.

A polynomial is factored completely when you have pulled out every possible common factor and cannot factor the remaining expressions using the patterns you know.

The central habit for this whole lesson is:

Always check for a greatest common factor first.

That remains true even when the expression also looks like a trinomial or a difference of squares.

A factoring decision chart: first remove any greatest common factor; then use the number of terms and recognizable patterns to choose a method. For this lesson, focus on the GCF, trinomial, and difference-of-squares branches.

The chart includes a few methods that will be developed later. For now, use this shorter decision process:

Find and factor out the GCF.
Look at what remains.
- With three terms, test for a trinomial pattern.
- With two terms and subtraction, test for a difference of squares.
Check whether the answer can be factored further.
Multiply to verify when needed.

1. Greatest common factor: reverse distribution

The greatest common factor, or GCF, is the largest expression that divides evenly into every term.

For a numerical GCF, find the greatest number dividing all coefficients. For variables, include only variables that occur in every term, using the smallest exponent present.

Consider:

12 x^{3} - 18 x^{2} + 6 x .

The coefficients $12$ , $18$ , and $6$ have a numerical GCF of $6$ . Each term contains $x$ , and the smallest power of $x$ is $x^{1}$ . Therefore, the full GCF is

6 x .

Divide every term by $6 x$ :

12 x^{3} \div 6 x = 2 x^{2}

- 18 x^{2} \div 6 x = - 3 x

6 x \div 6 x = 1.

So the first factoring step is

12 x^{3} - 18 x^{2} + 6 x = 6 x (2 x^{2} - 3 x + 1) .

Notice that the work is not finished: the trinomial inside the parentheses factors further.

2 x^{2} - 3 x + 1 = (2 x - 1) (x - 1) .

Thus the expression factored completely is

12 x^{3} - 18 x^{2} + 6 x = 6 x (2 x - 1) (x - 1) .

You can verify this by multiplying the last two binomials:

(2 x - 1) (x - 1) = 2 x^{2} - 3 x + 1.

Then distributing $6 x$ recreates the original polynomial.

How To Factor Polynomials The Easy Way!

Watch “How To Factor Polynomials The Easy Way!” from The Organic Chemistry Tutor for a concise demonstration of factoring out a numerical and variable GCF.

Watch GCF examples. Pay attention to the division used to determine each term inside the parentheses: every original term must be divided by the entire GCF, not just part of it.

Variables in the GCF

For an expression with more than one variable, treat each variable separately. Factor:

15 a^{3} b^{2} - 20 a^{2} b .

The numerical GCF of $15$ and $20$ is $5$ . Both terms contain $a$ , and the smaller exponent is $a^{2}$ . Both also contain $b$ , with smallest exponent $b^{1}$ . Therefore,

GCF = 5 a^{2} b .

Now divide each term by $5 a^{2} b$ :

15 a^{3} b^{2} \div 5 a^{2} b = 3 ab

- 20 a^{2} b \div 5 a^{2} b = - 4.

15 a^{3} b^{2} - 20 a^{2} b = 5 a^{2} b (3 ab - 4) .

Why factor out a negative GCF?

When the first term is negative, factoring out a negative common factor often makes the polynomial inside parentheses easier to work with.

- 8 y^{2} + 24 y - 16.

You could factor out $8$ , but that leaves a negative leading term inside:

8 (- y^{2} + 3 y - 2) .

Instead, factor out $- 8$ :

- 8 y^{2} + 24 y - 16 = - 8 (y^{2} - 3 y + 2) .

The trinomial inside now has a positive leading coefficient and factors more comfortably:

y^{2} - 3 y + 2 = (y - 1) (y - 2) .

Therefore,

- 8 y^{2} + 24 y - 16 = - 8 (y - 1) (y - 2) .

2. Factoring trinomials: undoing FOIL

A common quadratic trinomial has the form

x^{2} + b x + c .

To factor it, seek two integers whose:

product is $c$ , the constant term;
sum is $b$ , the coefficient of $x$ .

Why does this work? Suppose the factors have the form

(x + m) (x + n) .

Expanding gives

(x + m) (x + n) = x^{2} + (m + n) x + mn .

So the middle coefficient comes from $m + n$ , and the constant comes from $mn$ .

Example: unlike signs

Factor:

x^{2} - 2 x - 15.

We need two numbers whose product is $- 15$ and whose sum is $- 2$ .

The factor pairs of $- 15$ include:

Pair	Product	Sum
$1$ and $- 15$	$- 15$	$- 14$
$- 1$ and $15$	$- 15$	$14$
$3$ and $- 5$	$- 15$	$- 2$

The needed pair is $3$ and $- 5$ . Therefore,

x^{2} - 2 x - 15 = (x + 3) (x - 5) .

A quick expansion confirms it:

(x + 3) (x - 5) = x^{2} - 5 x + 3 x - 15 = x^{2} - 2 x - 15.

Sign strategy

The signs in a trinomial give useful information before you begin listing pairs.

Trinomial ending	Signs of the two numbers
$+ c$ and $+ b x$	Both positive
$+ c$ and $- b x$	Both negative
$- c$	One positive and one negative

For example, in

x^{2} + 7 x + 12,

the constant is positive and the middle coefficient is positive. Both numbers must be positive:

3 \cdot 4 = 12

and

3 + 4 = 7.

Therefore,

x^{2} + 7 x + 12 = (x + 3) (x + 4) .

How To Factor Polynomials The Easy Way!

Continue with The Organic Chemistry Tutor’s video to see the product-and-sum method, including cases where a GCF simplifies a trinomial first.

Watch monic trinomials for the basic product-and-sum method. Then watch GCF then factor to reinforce the rule that a common factor comes out before factoring the trinomial inside.

Perfect-square trinomials

Some trinomials come from squaring a binomial:

(A + B)^{2} = A^{2} + 2 A B + B^{2}

and

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

For example:

x^{2} + 10 x + 25.

The first term is a square:

x^{2} = (x)^{2},

and the last term is a square:

25 = 5^{2} .

Check the middle term:

2 (x) (5) = 10 x .

It matches, so the trinomial is a perfect square:

x^{2} + 10 x + 25 = (x + 5)^{2} .

Similarly,

x^{2} - 14 x + 49

has square end terms, and

2 (x) (7) = 14 x .

Because the middle term is negative, the binomial uses subtraction:

x^{2} - 14 x + 49 = (x - 7)^{2} .

Perfect-square recognition is a shortcut. You could still use the ordinary product-and-sum method, but noticing the pattern saves time and tells you that the two binomial factors are identical.

3. Trinomials with a leading coefficient other than 1

Now consider a trinomial such as

2 x^{2} - 5 x - 3.

This does not begin with $x^{2}$ , so the simple “two numbers multiply to $c$ ” method is not enough. We can use the AC method.

For a trinomial

a x^{2} + b x + c,

follow this process:

Multiply $a$ and $c$ .
Find two numbers whose product is $a c$ and whose sum is $b$ .
Split the middle term using those two numbers.
Factor the resulting four terms in pairs.
Factor out the common binomial.

For

2 x^{2} - 5 x - 3,

we have

a = 2, b = - 5, c = - 3.

First multiply $a$ and $c$ :

a c = 2 (- 3) = - 6.

We need two numbers with product $- 6$ and sum $- 5$ :

- 6 \cdot 1 = - 6

and

- 6 + 1 = - 5.

Split the middle term:

2 x^{2} - 5 x - 3 = 2 x^{2} - 6 x + x - 3.

Now factor the first two terms and the last two terms:

2 x^{2} - 6 x + x - 3

= 2 x (x - 3) + 1 (x - 3) .

Both groups contain $(x - 3)$ , so factor it out:

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

The multiplication check is especially valuable here:

(2 x + 1) (x - 3) = 2 x^{2} - 6 x + x - 3 = 2 x^{2} - 5 x - 3.

How To Factor Polynomials The Easy Way!

Use the next segment of The Organic Chemistry Tutor’s video to see the AC method and grouping carried out carefully on several non-monic trinomials.

Watch the AC method. Focus on how the middle term is split into two terms with the same total value, which makes a common binomial appear after factoring each pair.

If no pair of integers satisfies the required product and sum conditions, do not invent one. The trinomial may be prime with respect to integer factoring, or it may require methods beyond the scope of this lesson.

4. Difference of squares

A difference of squares has exactly two terms, joined by subtraction:

A^{2} - B^{2} .

It factors as

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

The factors are called conjugates because they are identical except for the sign between their terms.

Factor:

9 x^{2} - 49.

Both terms are perfect squares:

9 x^{2} = (3 x)^{2}

and

49 = 7^{2} .

Apply the pattern:

9 x^{2} - 49 = (3 x - 7) (3 x + 7) .

The reason this works is that the middle terms cancel when you multiply:

(3 x - 7) (3 x + 7)

= 9 x^{2} + 21 x - 21 x - 49

= 9 x^{2} - 49.

The cancellation is why one factor has a plus sign and the other has a minus sign.

How To Factor Polynomials The Easy Way!

Watch the difference-of-squares portion of The Organic Chemistry Tutor’s video to connect the pattern to square roots of the two terms.

Watch difference of squares. For each example, identify the square root of the first term and the square root of the second term before looking at the final factors.

Conditions that must all be true

Use the difference-of-squares pattern only when all three conditions hold:

There are exactly two terms.
The operation is subtraction.
Both terms are perfect squares.

For instance,

x^{2} + 25

is a sum of squares, not a difference, so it does not factor using this pattern over the real numbers.

Likewise,

x^{2} - 12

is a difference, but $12$ is not a perfect-square integer. It does not fit the standard integer difference-of-squares pattern.

Factor repeatedly when possible

Sometimes the difference-of-squares pattern appears only after the GCF is removed.

18 x^{2} - 72.

Start with the GCF:

18 x^{2} - 72 = 18 (x^{2} - 4) .

Then recognize the difference of squares:

x^{2} - 4 = x^{2} - 2^{2} .

Therefore,

18 x^{2} - 72 = 18 (x - 2) (x + 2) .

This is why stopping after the first factoring step can leave an answer incomplete.

A complete-factoring checklist

Before considering a factoring problem finished, use this checklist:

Check	What to look for
GCF first	Have you removed the greatest numerical and variable factor shared by every term?
Number of terms	Is the remaining expression a binomial or trinomial?
Pattern match	Is a trinomial factorable by product-and-sum or the AC method? Is a binomial a difference of squares?
Factor again	Can any factor still be broken down using the same patterns?
Verify	Does multiplying your factors reproduce the original polynomial?

For example, consider

6 x^{3} - 54 x .

The GCF is $6 x$ :

6 x^{3} - 54 x = 6 x (x^{2} - 9) .

The expression in parentheses is a difference of squares:

x^{2} - 9 = x^{2} - 3^{2} = (x - 3) (x + 3) .

So the fully factored result is

6 x^{3} - 54 x = 6 x (x - 3) (x + 3) .

Stopping at $6 x (x^{2} - 9)$ would be correct but not fully factored.

Key takeaways

Factoring reverses polynomial multiplication. The most dependable strategy is to inspect the polynomial in a consistent order:

Always factor out the GCF first. For variables, use the smallest exponent appearing in every term.
For a trinomial of the form $x^{2} + b x + c$ , find two numbers whose product is $c$ and whose sum is $b$ .
Recognize perfect-square trinomials such as

x^{2} + 10 x + 25 = (x + 5)^{2} .

For a trinomial $a x^{2} + b x + c$ with $a \neq = 1$ , the AC method organizes the factoring process.
A difference of squares has the form

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

Check for further factoring after every step, and use multiplication to verify your result.

Next, you will use factored polynomials to solve quadratic equations with the zero-product property.