Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 8 - Polynomials and Factoring - Chapter Review - 8-8 Factoring by Grouping: 76

Answer

$(11b - 6)(b^{2} + 1)$

Work Step by Step

$11b^{3} - 6b^{2} + 11b - 6$ Factor by grouping, group the first two terms and last two terms together. $(11b^{3} - 6b^{2}) + (11b - 6)$ Factor out the common factors from both parenthesis. $b^{2}(11b - 6) + 1(11b - 6)$ We take $(11b - 6)$ and factor it out which gives us. $(11b - 6)(b^{2} + 1)$
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