Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - Test: 18

Answer

$\color{blue}{(y-3)(y+3)(x-2)(x^2+2x+4)}$

Work Step by Step

Group the first two terms together and the last two terms together to obtain: $=(x^3y^2-9x^3)+(-8y^2+72)$ Factor out $x^3$ in the first group and $-8$ in the second group to obtain: $=x^3(y^2-9)+(-8)(y^2-9)$ Factor out the GCF $y^2-9$ to obtain: $=(y^2-9)(x^3-8) \\=(y^2-3^2)(x^3-2^3)$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $=(y-3)(y+3)(x^3-2^3)$ Factor the difference of two cubes using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x$ and $b=2$ to obtain: $=(y-3)(y+3)(x-2)(x^2+x(2)+2^2) \\=\color{blue}{(y-3)(y+3)(x-2)(x^2+2x+4)}$
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