Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.4 Factoring Polynomials - A.4 Assess Your Understanding - Page A40: 92


$$3y(y+2)(y-8) $$

Work Step by Step

We take out $3y$ from the given polynomial $$3y^3 - 18y^2 - 48y=3y(y^2 -6y -16)$$ To factor the remaining polynomial, we have to find integers whose product is $-16$ and whose sum is $-6$. That is, we have $2$ and $-8$, where $2-8=-6$ and $2\times (-8)=-16$. Thus $$ 3y^3 - 18y^2 - 48y=3y(y+2)(y-8) .$$
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