College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Test: 18

Answer

$(y+3)(y-3)(x-2)(x^2+2x+4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ x^3y^2-9x^3-8y^2+72 ,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, use factoring by grouping. This results to a factored expression that can still be factored using the difference of $2$ squares and the difference of $2$ cubes. Use the appropriate factoring method for these. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^3y^2-9x^3)-(8y^2-72) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^3(y^2-9)-8(y^2-9) .\end{array} Factoring the $GCF= (y^2-9) $ of the entire expression above results to \begin{array}{l}\require{cancel} (y^2-9)(x^3-8) .\end{array} The expressions $ y^2 $ and $ 9 $ are both perfect squares and are separated by a minus sign. Hence, $ y^2-9 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (y+3)(y-3)(x^3-8) .\end{array} The expressions $ x^3 $ and $ 8 $ are both perfect cubes (the cube root is exact). Hence, $ x^3-8 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent\begin{array}{l}\require{cancel} (y+3)(y-3)(x-2)(x^2+2x+4) .\end{array}
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