How to Factor the Difference of Two Perfect Squares

Reviewed by Joseph Meyer

Last Updated: April 8, 2024 Fact Checked

The difference of squares method is an easy way to factor a polynomial that involves the subtraction of two perfect squares. Using the formula $a^{{2}}-b^{{2}}=(a-b)(a+b)$ , you simply need to find the square root of each perfect square in the polynomial, and substitute those values into the formula. The difference of squares method is a basic tool in algebra that you will likely use often when solving equations.

Part 1

Part 1 of 3:

Evaluating the Polynomial

Download Article

1
Identify the coefficient, variable, and degree of each term. A coefficient is the number in front of a variable, which is multiplied by the variable. ^{[1]
X
Research source} The variable is the unknown value, usually denoted by $x$ or $y$ . ^{[2]
X
Research source} . The degree refers to the exponent of the variable. For example, a second-degree term has a value to the second power ( $x^{{2}}$ ) and a fourth-degree term has a value to the fourth power ( $x^{{4}}$ ). ^{[3]
X
Research source}
- For example, in the polynomial $36x^{{4}}-100x^{{2}}$ , the coefficients are $36$ and $100$ , the variable is $x$ , and the first term ( $36x^{{4}}$ ) is a fourth-degree term, and the second term ( $100x^{{2}}$ ) is a second-degree term.
2
Look for a greatest common factor. A greatest common factor is the highest factor that divides evenly into two or more terms. ^{[4]
X
Research source} If there is a factor common to both terms of the polynomial, factor this out. ^{[5]
X
Research source}
- For example, the two terms in the polynomial $36x^{{4}}-100x^{{2}}$ have a greatest common factor of $4x^{{2}}$ . Factoring this out, the problem becomes $4x^{{2}}(9x^{{2}}-25)$ .
Advertisement
3
Determine whether the terms are perfect squares. If you factored out a greatest common factor, you are only looking at the terms that remain inside the parentheses. A perfect square is the result of multiplying an integer by itself. ^{[6]
X
Research source} A variable is a perfect square if its exponent is an even number. You can only factor using the difference of squares if each term in the polynomial is a perfect square.
- For example, $9x^{{2}}$ is a perfect square, because $(3x)(3x)=9x^{{2}}$ . The number $25$ is also a perfect square, because $(5)(5)=25$ . Thus, you can factor $9x^{{2}}-25$ using the difference of squares formula.
4
Make sure you are finding the difference. You know you are finding the difference if you have a polynomial that subtracts one term from another. The difference of squares only applies to these polynomials, and not those in which addition is used.
- For example, you cannot factor $9x^{{2}}+25$ using the difference of squares formula, because in this polynomial you are finding a sum, not a difference.

Part 2

Part 2 of 3:

Using the Formula

Download Article

The formula is $a^{{2}}-b^{{2}}=(a-b)(a+b)$ . The terms $a^{{2}}$ and $b^{{2}}$ are the perfect squares in your polynomial, and $a$ and $b$ are the roots of the perfect squares. ^{[7]
X
Research source}
2
Plug the first term into the formula. This is the value for $a$ . To find this value, take the square root of the first perfect square in the polynomial. Remember that a square root of a number is a factor you multiply by itself to get that number. ^{[8]
X
Research source}
- For example, since $(3x)(3x)=9x^{{2}}$ , the square root of $9x^{{2}}$ is $3x$ . So you should substitute this value for $a$ in the difference of squares formula: $9x^{{2}}-25=(3x-b)(3x+b)$ .
3
Plug the second term into the formula. This is the value for $b$ , which is the square root of the second term in the polynomial. ^{[9]
X
Research source}
- For example, since $(5)(5)=25$ , the square root of $25$ is $5$ . So you should substitute this value for $b$ in the difference of squares formula: $9x^{{2}}-25=(3x-5)(3x+5)$ .
4
Check your work. Use the FOIL method to multiply the two factors. If your result is your original polynomial, you know you have factored correctly. ^{[10]
X
Research source}
- For example:
  $(3x-5)(3x+5)$
  $=9x^{{2}}+15x-15x-25$
  $=9x^{{2}}-25$ .

Part 3

Part 3 of 3:

Solving Practice Problems

Download Article

1
Factor this polynomial. Use the difference of two squares formula: $36x^{{4}}-9$ .
- The terms have no greatest common factor, so there is no need to factor anything out of the polynomial.
- The term $36x^{{4}}$ is a perfect square, since $(6x^{{2}})(6x^{{2}})=36x^{{4}}$ .
- The term $9$ is a perfect square, since $(3)(3)=9$ .
- The difference of squares formula is $a^{{2}}-b^{{2}}=(a-b)(a+b)$ . Thus, $36x^{{4}}-9=(a-b)(a+b)$ , where $a$ and $b$ are the square roots of the perfect squares.
- The square root of $36x^{{4}}$ is $6x^{{2}}$ . Plugging in for $a$ you have $36x^{{4}}-9=(6x^{{2}}-b)(6x^{{2}}+b)$ .
- The square root of $9$ is $3$ . So plugging in for $b$ , you have $36x^{{4}}-9=(6x^{{2}}-3)(6x^{{2}}+3)$ .
2
Try factoring this polynomial. Make sure you factor out a greatest common factor, and use the difference of two squares: $48x^{{3}}-27x$ .
- Find the greatest common factor of each term. This term is $3x$ , so factor this out of the polynomial: $3x(16x^{{2}}-9)$ .
- The term $16x^{{2}}$ is a perfect square, since $(4x)(4x)=16x^{{2}}$ .
- The term $9$ is a perfect square, since $(3)(3)=9$ .
- The difference of squares formula is $a^{{2}}-b^{{2}}=(a-b)(a+b)$ . Thus, $48x^{{3}}-27x=3x(a-b)(a+b)$ , where $a$ and $b$ are the square roots of the perfect squares.
- The square root of $16x^{{2}}$ is $4x$ . Plugging in for $a$ you have $48x^{{3}}-27x=3x(4x-b)(4x+b)$ .
- The square root of $9$ is $3$ . So plugging in for $b$ , you have $48x^{{3}}-27x=3x(4x-3)(4x+3)$ .
3
Factor the following polynomial. It has two variables, but it still follows the rules for the difference of squares method: $4x^{{2}}-81y^{{2}}$ .
- No factor is common to each term in this polynomial, so there is nothing to factor out before you begin factoring the difference of squares.
- The term $4x^{{2}}$ is a perfect square, since $(2x)(2x)=4x^{{2}}$ .
- The term $81y^{{2}}$ is a perfect square, since $(9y)(9y)=81y^{{2}}$ .
- The difference of squares formula is $a^{{2}}-b^{{2}}=(a-b)(a+b)$ . Thus, $4x^{{2}}-81y^{{2}}=(a-b)(a+b)$ , where $a$ and $b$ are the square roots of the perfect squares.
- The square root of $4x^{{2}}$ is $2x$ . Plugging in for $a$ you have $4x^{{2}}-81y^{{2}}=(2x-b)(2x+b)$ .
- The square root of $81y^{{2}}$ is $9y$ . So plugging in for $b$ , you have $4x^{{2}}-81y^{{2}}=(2x-9y)(2x+9y)$ .