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1.15: Factoring the Difference of Two Squares

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    45448

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    In this chapter, we will learn how to factor a binomial that is a difference of two perfect squares. We have learned in multiplying polynomials that a product of two conjugates yields a difference of two perfect squares:

    \[(a+b)(a-b)=a^{2}-a b+a b-b^{2}=a^{2}-b^{2}\nonumber\]

    This indicates that the factor form of \(a^{2}-b^{2}\) is \((a+b)(a-b),\) a product of two conjugates. Let's put this as a formula:

    Factoring the Difference of Two Squares

    \[a^{2}-b^{2}=(a+b)(a-b)\nonumber\]

    Example 13.1

    Factor a difference of two squares.

    1. \(49-y^{2}=(7)^{2}-y^{2}=(7+y)(7-y)\)
    2. \(16 w^{2}-x^{2} y^{2}=(4 w)^{2}-(x y)^{2}=(4 w+x y)(4 w-x y)\)
    3. \(9 a^{6}-b^{4}=\left(3 a^{3}\right)^{2}-\left(b^{2}\right)^{2}=\left(3 a^{3}+b^{2}\right)\left(3 a^{3}-b^{2}\right)\)

    Sometimes, the binomial is not a difference of two perfect squares, but after we factor out the GCF, the resulting binomial is a difference of two perfect squares. Then we can still use this formula to continue factoring the resulting binomial.

    Example 13.2

    Factor the binomial completely.

    1. \(18 x^{3}-8 x y^{2}=2 x\left(9 x^{2}-4 y^{2}\right)=2 x\left[(3 x)^{2}-(2 y)^{2}\right]=2 x(3 x+2 y)(3 x-2 y)\)
    2. \(3 a^{5}-27 a b^{2}=3 a\left(a^{4}-9 b^{2}\right)=3 a\left[\left(a^{2}\right)^{2}-(3 b)^{2}\right]=3 a\left(a^{2}+3 b\right)\left(a^{2}-3 b\right)\)

    Exit Problem

    Factor completely: \(16 x^{2}-36\)

    This page titled 1.15: Factoring the Difference of Two Squares is shared under a CC BY-NC-ND 4.0 license and was authored, remixed, and/or curated by Samar ElHitti, Marianna Bonanome, Holly Carley, Thomas Tradler, & Lin Zhou (New York City College of Technology at CUNY Academic Works) .