Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 4

Answer

$(3a-5)^2$

Work Step by Step

The two numbers whose product is $ac= 9(25)=225 $ and whose sum is $b= -30 $ are $\{ -15,-15 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 9a^2-30a+25 $, is \begin{array}{l}\require{cancel} 9a^2-15a-15a+25 \\\\= (9a^2-15a)-(15a-25) \\\\= 3a(3a-5)-5(3a-5) \\\\= (3a-5)(3a-5) \\\\= (3a-5)^2 .\end{array}
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