Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 34: 122

Answer

$9x^{3}+18x^{2}-x-2=(x+2)(3x-1)(3x+1)$

Work Step by Step

$9x^{3}+18x^{2}-x-2$ Group the first and third terms together. Do the same with the second and fourth terms: $9x^{3}+18x^{2}-x-2=(9x^{3}-x)+(18x^{2}-2)=...$ Take out common factor $x$ from the first parentheses and common factor $2$ from the second parentheses: $...=x(9x^{2}-1)+2(9x^{2}-1)=...$ Take out common factor $(9x^{2}-1)$: $...=(x+2)(9x^{2}-1)=...$ Finally, factor the difference of squares in the second parentheses: $...=(x+2)(3x-1)(3x+1)$
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