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How to Factor the Difference of Two Perfect Squares

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Reviewed by Joseph Meyer

Last Updated: April 8, 2024 Fact Checked

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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]
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.
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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]
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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 $4x^{{2}}(9x^{{2}}-25)$ .
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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]
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.
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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.

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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]
X
Research source}
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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: $9x^{{2}}-25=(3x-b)(3x+b)$ .
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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: $9x^{{2}}-25=(3x-5)(3x+5)$ .
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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:
  $(3x-5)(3x+5)$
  $=9x^{{2}}+15x-15x-25$
  $=9x^{{2}}-25$ .

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Part 3

Part 3 of 3:

Solving Practice Problems

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1
Factor this polynomial. Use the difference of two squares formula: $36x^{{4}}-9$ .
- 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 $36x^{{4}}-9=(6x^{{2}}-3)(6x^{{2}}+3)$ .
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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 $48x^{{3}}-27x=3x(4x-3)(4x+3)$ .
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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 $4x^{{2}}-81y^{{2}}=(2x-9y)(2x+9y)$ .

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References

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↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm

About This Article

Reviewed by:

Math Teacher

This article was reviewed by Joseph Meyer . Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. This article has been viewed 51,699 times.

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Co-authors: 9

Updated: April 8, 2024

Views: 51,699

Categories: Algebra

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