College Algebra (10th Edition)

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

Chapter R - Section R.8 - nth Roots; Rational Exponents - R.8 Assess Your Understanding - Page 80: 107

Answer

$(3x+5)^{1/3}(2x+3)^{1/2}(17x+27)$

Work Step by Step

We simplify: $4(3x+5)^{1/3}(2x+3)^{3/2}+3(3x+5)^{4/3}(2x+3)^{1/2} =(3x+5)^{1/3}(2x+3)^{1/2}[4(2x+3)^{2/2}+3(3x+5)^{3/3}] =(3x+5)^{1/3}(2x+3)^{1/2}[4(2x+3)+3(3x+5)] =(3x+5)^{1/3}(2x+3)^{1/2}(8x+12+9x+15) =(3x+5)^{1/3}(2x+3)^{1/2}(17x+27)$
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