Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 30



Work Step by Step

Let $z=(4r+1)$. Then the given expression, $ (4r+1)^2+8(4r+1)+16 $, is equivalent to \begin{array}{l} z^2+8z+16 .\end{array} The two numbers whose product is $ac= 1(16)=16 $ and whose sum is $b= 8 $ are $\{ 4,4 \}$. Using these two numbers to decompose the middle term of the expression, $ z^2+8z+16 $, results to \begin{array}{l} z^2+4z+4z+16 \\\\= (z^2+4z)+(4z+16) \\\\= z(z+4)+4(z+4) \\\\= (z+4)(z+4) \\\\= (z+4)^2 .\end{array} Since $z=(4r+1)$, then, \begin{array}{l} (z+4)^2 \\\\= (4r+1+4)^2 \\\\= (4r+5)^2 .\end{array}
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