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: 107


$$(2y-3)(2y-5) .$$

Work Step by Step

First, for the polynomial $4y^2 - 16y + 15$, we write $$4y^2 - 16y + 15=4y^2 - 6y-10y + 15 .$$ By grouping the first two terms and the second two terms and then taking common factors, we have $$4y^2 - 6y-10y + 15=2y(2y-3)-5(2y-3)=(2y-3)(2y-5) .$$
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