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

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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]
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Research source} The variable is the unknown value, usually denoted by $x$ $x$ or $y$ $y$ .^{[2]
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Research source}. The degree refers to the exponent of the variable. For example, a second-degree term has a value to the second power ( $x^{2}$ $x^{{2}}$ ) and a fourth-degree term has a value to the fourth power ( $x^{4}$ $x^{{4}}$ ).^{[3]
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- 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]
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Research source} If there is a factor common to both terms of the polynomial, factor this out.^{[5]
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- 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 $4x^{2}(9x^{2}-25)$ .
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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]
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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

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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]
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2
Plug the first term into the formula. This is the value for $a$ $a$ . 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]
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- 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 $b$ $b$ , which is the square root of the second term in the polynomial.^{[9]
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- 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]
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- For example:
  $(3x-5)(3x+5)$
  $=9x^{2}+15x-15x-25$
  $=9x^{2}-25$ .

Part 3

Part 3 of 3:

Solving Practice Problems

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1
Factor this polynomial. Use the difference of two squares formula: $36x^{4}-9$ $36x^{{4}}-9$ .
- 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: $48x^{3}-27x$ $48x^{{3}}-27x$ .
- 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: $4x^{2}-81y^{2}$ $4x^{{2}}-81y^{{2}}$ .
- 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 $4x^{2}-81y^{2}=(2x-9y)(2x+9y)$ .