College Algebra (11th Edition)

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

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises - Page 40: 102

Answer

$(3a-4)^2$

Work Step by Step

Let $z=(3a+5).$ Then the given expression, $ (3a+5)^2-18(3a+5)+81 ,$ is equivalent to \begin{array}{l}\require{cancel} z^2-18z+81 .\end{array} The two numbers whose product is $ac= 1(81)=81 $ and whose sum is $b= -18 $ are $\{ -9,-9 \}$. Using these two numbers to decompose the middle term of the expression, $ z^2-18z+81 ,$ then \begin{array}{l}\require{cancel} z^2-9z-9z+81 \\\\= (z^2-9z)-(9z-81) \\\\= z(z-9)-9(z-9) \\\\= (z-9)(z-9) \\\\= (z-9)^2 .\end{array} Since $z=(3a+5)$, then the expression $ (z-9)^2 $ is equivalent to \begin{array}{l}\require{cancel} (3a+5-9)^2 \\\\= (3a-4)^2 .\end{array}
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