Intermediate Algebra (6th Edition)

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

Chapter 10 - Cumulative Review - Page 633: 10b

Answer

$2a^3(5a+1)(2a+5)$

Work Step by Step

Factoring the $GCF=2a^3$ results to $ 2a^3(10a^2+27a+5) $. The 2 numbers whose product is $ac= 10(5)=50 $ and whose sum is $b= 27 $ are $\left\{ 2, 25 \right\}$. Using these numbers to decompose the middle term results to \begin{array}{l} 2a^3(10a^2+2a+25a+5) \\\\= 2a^3[(10a^2+2a)+(25a+5)] \\\\= 2a^3[2a(5a+1)+5(5a+1)] \\\\= 2a^3[(5a+1)(2a+5)] \\\\= 2a^3(5a+1)(2a+5) .\end{array}
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