Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 28

Answer

$(3a-4+b)(3a-4-b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 9a^2-24a+16-b^2 ,$ group the first $3$ terms since these form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms above results to \begin{array}{l}\require{cancel} (9a^2-24a+16)-b^2 .\end{array} The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (3a-4)^2-b^2 .\end{array} The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is \begin{array}{l}\require{cancel} [(3a-4)+b][(3a-4)-b] \\\\= (3a-4+b)(3a-4-b) .\end{array}
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