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: 95



Work Step by Step

We have $6x^2 + 8x + 2=2(3x^2+4x+1)=2(3x^2+3x+x+1)$ By grouping the first two terms and the second two terms, and then taking common factors, we have $$ 2(3x^2+3x+x+1)=2(3x(x+1)+(x+1))=2(x+1)(3x+1) .$$
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