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: 29



Work Step by Step

Let $z=(5a-3)$. Then the given expression, $ (5a-3)^2-6(5a-3)+9 $, is equivalent to \begin{array}{l} z^2-6z+9 .\end{array} The two numbers whose product is $ac= 1(9)=9 $ and whose sum is $b= -6 $ are $\{ -3,-3 \}$. Using these two numbers to decompose the middle term of the expression, $ z^2-6z+9 $, results to \begin{array}{l} z^2-3z-3z+9 \\\\= (z^2-3z)-(3z-9) \\\\= z(z-3)-3(z-3) \\\\= (z-3)(z-3) \\\\= (z-3)^2 .\end{array} Since $z=(5a-3)$, then, \begin{array}{l} (z-3)^2 \\\\= (5a-3-3)^2 \\\\= (5a-6)^2 .\end{array}
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