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: 55



Work Step by Step

Let $z=(y-6)$. Then the given expression, $ (y-6)^2+3(y-6)+2 $, is equivalent to \begin{array}{l} z^2+3z+2 .\end{array} The two numbers whose product is $ac= 1(2)=2 $ and whose sum is $b= 3 $ are $\{ 1,2 \}$. Using these two numbers to decompose the middle term of the expression, $ z^2+3z+2 $, results to \begin{array}{l} z^2+1z+2z+2 \\\\= z(z+1)+2(z+1) \\\\= (z+1)(z+2) .\end{array} Since $z=(y-6)$, then, \begin{array}{l} (z+1)(z+2) \\\\= (y-6+1)(y-6+2) \\\\= (y-5)(y-4) .\end{array}
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