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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 , 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. A coefficient is the number in front of a variable, which is multiplied by the variable. [1] The variable is the unknown value, usually denoted by or . [2] . The degree refers to the exponent of the variable. For example, a second-degree term has a value to the second power ( ) and a fourth-degree term has a value to the fourth power ( ). [3]
    • For example, in the polynomial , the coefficients are and , the variable is , and the first term ( ) is a fourth-degree term, and the second term ( ) is a second-degree term.
  2. A greatest common factor is the highest factor that divides evenly into two or more terms. [4] If there is a factor common to both terms of the polynomial, factor this out. [5]
    • For example, the two terms in the polynomial have a greatest common factor of . Factoring this out, the problem becomes .
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  3. 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] 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, is a perfect square, because . The number is also a perfect square, because . Thus, you can factor using the difference of squares formula.
  4. 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 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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  1. The formula is . The terms and are the perfect squares in your polynomial, and and are the roots of the perfect squares. [7]
  2. This is the value for . 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]
    • For example, since , the square root of is . So you should substitute this value for in the difference of squares formula: .
  3. This is the value for , which is the square root of the second term in the polynomial. [9]
    • For example, since , the square root of is . So you should substitute this value for in the difference of squares formula: .
  4. Use the FOIL method to multiply the two factors. If your result is your original polynomial, you know you have factored correctly. [10]
    • For example:


      .
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Part 3
Part 3 of 3:

Solving Practice Problems

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  1. Use the difference of two squares formula: .
    • The terms have no greatest common factor, so there is no need to factor anything out of the polynomial.
    • The term is a perfect square, since .
    • The term is a perfect square, since .
    • The difference of squares formula is . Thus, , where and are the square roots of the perfect squares.
    • The square root of is . Plugging in for you have .
    • The square root of is . So plugging in for , you have .
  2. Make sure you factor out a greatest common factor, and use the difference of two squares: .
    • Find the greatest common factor of each term. This term is , so factor this out of the polynomial: .
    • The term is a perfect square, since .
    • The term is a perfect square, since .
    • The difference of squares formula is . Thus, , where and are the square roots of the perfect squares.
    • The square root of is . Plugging in for you have .
    • The square root of is . So plugging in for , you have .
  3. It has two variables, but it still follows the rules for the difference of squares method: .
    • 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 is a perfect square, since .
    • The term is a perfect square, since .
    • The difference of squares formula is . Thus, , where and are the square roots of the perfect squares.
    • The square root of is . Plugging in for you have .
    • The square root of is . So plugging in for , you have .
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