College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.5 - Factoring Polynomials - R.5 Exercises - Page 58: 135

Answer

$6(3x-5)(2x+1)^{2}(5x-4)$

Work Step by Step

We factor by grouping to obtain: $2(3x-5)* 3(2x+1)^{3}+(3x-5)^{2}* 3(2x+1)^{2}* 2 =6(3x-5)(2x+1)^{3}+6(3x-5)^{2}(2x+1)^{2} =6(3x-5)(2x+1)^{2}[(2x+1)^1+(3x-5)^1] =6(3x-5)(2x+1)^{2}(2x+1+3x-5) =6(3x-5)(2x+1)^{2}(5x-4)$
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