Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 687: 103



Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} w^3-4w+3w^2-12 \\\\= (w^3-4w)+(3w^2-12) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} w(w^2-4)+3(w^2-4) .\end{array} Factoring the $GCF= (w^2-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (w^2-4)(w+3) .\end{array} The expressions $ w^2 $ and $ 4 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ w^2-4 ,$ 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} [(w)^2-(2)^2](w+3) \\\\= [(w+2)(w-2)](w+3) \\\\= (w+2)(w-2)(w+3) .\end{array}
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