Factoring Binomials (Difference of Squares)

Q: Why do the middle terms cancel when I multiply $(a + b)(a - b)$?

FOIL it: First gives $a^{2}$, Outer gives $-ab$, Inner gives $+ab$, Last gives $-b^{2}$. The Outer and Inner terms are opposites, so they add to zero — leaving $a^{2} - b^{2}$ with no middle term.

A difference of squares is a two-term polynomial where both terms are perfect squares and they are being subtracted: $x^{2} - 16$ , $9x^{2} - 25$ , $4x^{2} - 49$ . Every one of them factors the same way: $a^{2} - b^{2} = (a + b)(a - b)$ .

The pattern works because multiplying $(a + b)(a - b)$ makes the middle terms cancel — Outer gives $-ab$ and Inner gives $+ab$ , and they wipe each other out, leaving just $a^{2} - b^{2}$ . Factoring runs that product in reverse: spot the two squares, write one plus and one minus.

Spotting the pattern

Three things must be true: exactly two terms, a subtraction between them, and both terms perfect squares. In $x^{2} - 16$ , the first term is $x^{2}$ (the square of $x$ ) and $16 = 4^{2}$ , so it qualifies with $a = x$ and $b = 4$ .

Coefficients count too. $9x^{2}$ is a perfect square because $9x^{2} = (3x)^{2}$ — take the square root of the coefficient and halve the exponent. So $9x^{2} - 25$ factors with $a = 3x$ and $b = 5$ .

Once you have $a$ and $b$ , the answer writes itself: $(a + b)(a - b)$ . The order of the two factors does not matter.

A sum of squares does not factor

The subtraction is not optional. $x^{2} + 36$ is a sum of squares, and it does not factor over the real numbers. Try it: $(x + 6)(x - 6) = x^{2} - 36$ , and $(x + 6)(x + 6) = x^{2} + 12x + 36$ . No pair of binomials produces $x^{2} + 36$ , because any middle terms either cancel (giving a difference) or survive (giving a trinomial).

On a test, "cannot be factored" is the correct answer for a sum of squares — writing $(x + 6)(x - 6)$ for $x^{2} + 36$ is the trap the question is checking for.

Check for a GCF first

Some binomials hide the pattern behind a common factor. $4x^{2} - 36$ is technically a difference of squares as written, but factoring the GCF of $4$ first gives $4(x^{2} - 9)$ , and then the difference of squares finishes it: $4(x + 3)(x - 3)$ .

GCF first, pattern second — that order keeps the numbers small and guarantees a completely factored answer.

Worked examples

Example 1: the basic pattern

Factor $x^{2} - 49$ .

Identify the squares

x^{2} = (x)^{2}, \quad 49 = 7^{2}

Apply the pattern with

a = x

b = 7

a^{2} - b^{2} = (a + b)(a - b)

Write the factors

(x + 7)(x - 7)

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

Example 2: a coefficient on the squared term

Factor $4x^{2} - 25$ .

Identify the squares

4x^{2} = (2x)^{2}, \quad 25 = 5^{2}

Apply the pattern with

a = 2x

b = 5

(2x + 5)(2x - 5)

Answer: $(2x + 5)(2x - 5)$

Example 3: GCF first

Factor $4x^{2} - 36$ completely.

Factor out the GCF of

4

4(x^{2} - 9)

Identify the squares inside

x^{2} = (x)^{2}, \quad 9 = 3^{2}

Apply the pattern

4(x + 3)(x - 3)

Answer: $4(x + 3)(x - 3)$

Try one yourself

Common questions

Why do the middle terms cancel when I multiply

(a + b)(a - b)

FOIL it: First gives $a^{2}$ , Outer gives $-ab$ , Inner gives $+ab$ , Last gives $-b^{2}$ . The Outer and Inner terms are opposites, so they add to zero — leaving $a^{2} - b^{2}$ with no middle term.

Can

x^{2} + 36

be factored some other way?

Not with real numbers. A sum of squares has no real factoring — the check is quick: no two real binomials multiply to it. (In Algebra 2 you will factor it with imaginary numbers, but for now the answer is that it does not factor.)

How do I know if a term like

9x^{2}

is a perfect square?

Ask two questions: is the coefficient a perfect square, and is the exponent even? $9x^{2} = (3x)^{2}$ passes both. Something like $8x^{2}$ fails, because $8$ is not a perfect square.

Does the order

(a + b)(a - b)

versus

(a - b)(a + b)

matter?

No. Multiplication is commutative, so both orders are the same answer. What matters is that one factor has a plus and the other has a minus between the same two terms.

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Factoring Binomials (Difference of Squares)

Spotting the pattern

A sum of squares does not factor

Check for a GCF first

Worked examples

Example 1: the basic pattern

Example 2: a coefficient on the squared term

Example 3: GCF first

Try one yourself

Common questions

Keep learning

Factoring Binomials (Perfect Squares)

Square of a Binomial

Factoring by GCF

Difference of Squares