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: 83

Answer

$4(x^2-2x+8)$

Work Step by Step

To factor the second-degree polynomial $4x^2 - 8x + 32=4(x^2-2x+8)$, we have to find integers whose product is $8$ and whose sum is $-2$. But there are no such factors. Thus, the polynomial $x^2-2x+8$ is prime and hence we have $$ 4x^2 - 8x + 32=4(x^2-2x+8) .$$
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