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: 132



Work Step by Step

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