Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.4 - Factoring Trinomials - Exercise Set - Page 361: 74

Answer

$y^{3}(2y-3)(5y-1)$

Work Step by Step

$ 10y^{5}-17y^{4}+3y^{3}\qquad$...factor out the common term, $y^{3}$. $=y^{3}(10y^{2}-17y+3)$ Two factors of $ac=30$, whose sum is $b=-17$ are$ -15$ and $-2$. Rewrite the middle term and factor in pairs. $y^{3}(10y^{2}-17y+3)$ $=y^{3}(10y^{2}-15y-2y+3)$ $=y^{3}[5y(2y-3)-(2y-3)]$ =$y^{3}(2y-3)(5y-1)$
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