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: 73

Answer

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

Work Step by Step

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