How do I know when to use Difference of Squares instead of other factoring methods?

Difference of Squares is your tool when you can see a binomial where both terms are perfect squares and they are being subtracted. If you do not see that pattern clearly, or if there are three or more terms, reach for Factoring Trinomials or grouping instead. The key is recognizing both terms as perfect squares with a minus sign between them.

Why do we call it 'Difference of Squares' and not just 'factoring binomials'?

The name comes from what the expression is: a difference (subtraction) of two perfect squares. The word Difference tells you the operation is subtraction, and Squares tells you both terms are perfect squares. This specific pattern has a special factoring rule, unlike other binomials.

Does Difference of Squares work if I have a^4 - b^4?

Yes, if you recognize that a^4 = (a^2)^2 and b^4 = (b^2)^2, you can factor in stages: first a^4 - b^4 = (a^2 + b^2)(a^2 - b^2), then apply the pattern again to get (a^2 + b^2)(a + b)(a - b). This works because both a^4 and b^4 are still perfect squares.

What is the difference between Difference of Squares and Sum of Squares?

Difference of Squares (a^2 - b^2) factors neatly into (a + b)(a - b) using a simple formula. Sum of Squares (a^2 + b^2) does not factor using real numbers—it is a dead end without complex numbers. The minus sign is what makes the formula work, so always check your operation first.

Difference of Squares

Instantly factor binomials that are differences of perfect squares by recognizing the pattern and splitting into a product of a sum and difference.

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

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Type the problem. The solver will use Difference of Squares where Difference of Squares is the right tool, and tell you when it is not.

What each symbol means

What Difference of Squares takes

Symbol	Meaning
$a$	$a$ is the base of the first perfect square in the pattern; it represents any real number, expression, or variable, and changing its sign changes the factors.
$b$	$b$ is the base of the second perfect square; it can be any real number, expression, or variable, and the formula fails if you misidentify what is being squared.

When to use it

You reach for Difference of Squares when you see a binomial where both terms are perfect squares connected by a minus sign.

Level

Usually taught in: Algebra I · Appears on: SAT

Worked examples

1. Factor a simple difference of perfect squares

Problem

Factor

x^2 - 9

$x^2 - 9 = x^2 - 3^2$
Rewrite $9$ as $3^2$ so the pattern $a^2 - b^2$ becomes clear, with $a = x$ and $b = 3$ .
$x^2 - 3^2 = (x + 3)(x - 3)$
Apply $a^2 - b^2 = (a + b)(a - b)$ with $a = x$ and $b = 3$ to factor the expression.
$(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9$
Expand to verify: the middle terms cancel, leaving $x^2 - 9$ , confirming our factorization.

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

This is the simplest application of Difference of Squares: both $x^2$ and $9$ are obviously perfect squares, so we immediately recognize the pattern and apply the formula.

2. Factor with coefficients and higher exponents

Problem

Factor

4x^2 - 25y^4

$4x^2 - 25y^4 = (2x)^2 - (5y^2)^2$
Rewrite each term as a perfect square: $4x^2 = (2x)^2$ because $(2x)^2 = 4x^2$ , and $25y^4 = (5y^2)^2$ because $(5y^2)^2 = 25y^4$ .
$(2x)^2 - (5y^2)^2 = (2x + 5y^2)(2x - 5y^2)$
Apply $a^2 - b^2 = (a + b)(a - b)$ with $a = 2x$ and $b = 5y^2$ .
$(2x + 5y^2)(2x - 5y^2) = 4x^2 - 10xy^2 + 10xy^2 - 25y^4 = 4x^2 - 25y^4$
Expand to confirm: the middle terms cancel, giving us back the original expression.

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

This example requires recognizing that $4x^2$ and $25y^4$ are perfect squares even though they have coefficients and exponents; the method still works once you identify $a = 2x$ and $b = 5y^2$ .

3. Find remaining area at a bake sale

Problem

At a bake sale, a volunteer sets up a large square table with side length

d

feet to display cookies. A smaller square donation box with side length

1

foot is placed at the center. Factor the expression for the available display area.

$d^2 - 1$
The table has area $d^2$ square feet, and the donation box occupies $1$ square foot.
$d^2 - 1^2$
We write $1 = 1^2$ to see that both terms are perfect squares.
$d^2 - 1^2 = (d + 1)(d - 1)$
Apply the Difference of Squares formula with $a = d$ and $b = 1$ to factor completely.

Answer: $(d + 1)(d - 1) \text{ square feet}$

This real-world example shows how Difference of Squares gives us a factored form that makes it easier to understand or calculate the remaining space based on the table's dimensions.

Common mistakes

Where Difference of Squares usually goes wrong

The mistake	Why it is wrong	The fix
Writing $a^2 - b^2 = (a - b)(b - a)$ or $a^2 - b^2 = (a - b)^2$	Both are wrong. The first expression gives the negative of what we want, and the second is the formula for a perfect square trinomial—it expands with a middle term and does not equal the original difference.	Remember that $a^2 - b^2 = (a + b)(a - b)$ : one factor always adds the bases, one subtracts. Using $(a - b)^2$ gives $a^2 - 2ab + b^2$ , which includes a middle term not in $a^2 - b^2$ .
Trying to apply the formula to $a^2 + b^2$ instead of $a^2 - b^2$	Difference of Squares only works when the operation is subtraction, not addition; the sum of two squares does not factor using this pattern over the real numbers.	Check your operation first. For $x^2 + 25$ , Difference of Squares does not apply—look for a different method instead, such as recognizing that sum of squares has no real factorization.
Forgetting to rewrite terms with coefficients as perfect squares, such as treating $4x^2 - 9$ directly as $(4x + 3)(4x - 3)$	This gives the wrong factors because you did not first identify the bases correctly; expanding $(4x + 3)(4x - 3)$ does not give $4x^2 - 9$ .	Always rewrite coefficient-heavy terms first: $4x^2 = (2x)^2$ and $9 = 3^2$ , so $4x^2 - 9 = (2x)^2 - 3^2 = (2x + 3)(2x - 3)$ .

Tips and when to use something else

Always check that you have a minus sign, not a plus, between the two terms—Difference of Squares only works for subtraction.
If the terms have coefficients or exponents, rewrite them as perfect squares first: $9x^4 - 25y^2 = (3x^2)^2 - (5y)^2$ before applying the formula.
When neither term is obviously a perfect square, do not force it—look for a different factoring method like Factoring Trinomials or grouping instead.
Sum of squares — like $a^2 + b^2$ — does not factor over the reals, so Difference of Squares will not help; use a different technique for problems with a plus sign.

Related formulas

Perfect Square TrinomialQuickly factor trinomials of the form a^2 \pm 2ab + b^2 by recognizing the perfect square pattern that gives (a \pm b)^2.Factoring TrinomialsBreak a trinomial into two binomials by finding two numbers that add to b and multiply to c; use when solving equations or simplifying.Difference of CubesDifference of Cubes is a factoring pattern that breaks down any cubic binomial subtraction into a linear and a quadratic factor.Greatest Common Factor of a PolynomialFactor a polynomial by identifying and pulling out the single largest expression that divides evenly into every term in the polynomial.

Frequently asked questions

How do I know when to use Difference of Squares instead of other factoring methods?: Difference of Squares is your tool when you can see a binomial where both terms are perfect squares and they are being subtracted. If you do not see that pattern clearly, or if there are three or more terms, reach for Factoring Trinomials or grouping instead. The key is recognizing both terms as perfect squares with a minus sign between them.
Why do we call it 'Difference of Squares' and not just 'factoring binomials'?: The name comes from what the expression is: a difference (subtraction) of two perfect squares. The word Difference tells you the operation is subtraction, and Squares tells you both terms are perfect squares. This specific pattern has a special factoring rule, unlike other binomials.
Does Difference of Squares work if I have $a^4 - b^4$ ?: Yes, if you recognize that $a^4 = (a^2)^2$ and $b^4 = (b^2)^2$ , you can factor in stages: first $a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$ , then apply the pattern again to get $(a^2 + b^2)(a + b)(a - b)$ . This works because both $a^4$ and $b^4$ are still perfect squares.
What is the difference between Difference of Squares and Sum of Squares?: Difference of Squares ( $a^2 - b^2$ ) factors neatly into $(a + b)(a - b)$ using a simple formula. Sum of Squares ( $a^2 + b^2$ ) does not factor using real numbers—it is a dead end without complex numbers. The minus sign is what makes the formula work, so always check your operation first.

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Reviewed 2026-09-18