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: 130

Answer

$x^{2}(32x-9)$

Work Step by Step

We factor by grouping to obtain: $3x^{2}(8x-3)+x^{3}* 8 =3x^{2}(8x-3)+8x*x^2 =x^{2}(3(8x-3)+8x) =x^{2}(24x-9+8x) =x^{2}(32x-9)$
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