Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 304: 86

Answer

$3x^2y(y+5)(y-3)$

Work Step by Step

Factoring the $GCF= 3x^2y $ results to $ 3x^2y(y^2+2y-15) $. The two numbers whose product is $ -15 $ and whose sum is $ 2 $ are $\{ 5,-3 \}.$ Using these numbers to decompose the middle term of the trinomial results to \begin{align*} 3x^2y(y^2+2y-15) \\\Rightarrow 3x^2y(y^2+5y-3y-15) \\= 3x^2y[(y^2+5y)-(3y+15)] \\= 3x^2y[y(y+5)-3(y+5)] \\= 3x^2y[(y+5)(y-3)] \\= 3x^2y(y+5)(y-3) .\end{align*}
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