College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.5 - Page 75: 102

Answer

$\frac{2 (100x^{2} + 95x + 11)}{(4x + 3)^{2}}$

Work Step by Step

$-8(4x+3)^{-2}+ 10(5x+1)(4x+3)^{-1}$ The greatest common factor is $(4x+3)$ with the smaller exponent in the two terms. Thus the greatest common factor is $(4x+3)^{-2}$ Express each term as the product of greatest common factor and its other factor. $=-8(4x+3)^{-2}+ 10(5x+1)(4x+3)^{-2}(4x+3)$ $[(4x+3)^{-2}(4x+3) = (4x+3)^{-2+1} = (4x+3)^{-1} ]$ Factor out the Greatest common factor. $=(4x+3)^{-2}(-8+10(5x+1)(4x+3))$ Multiply (5x+1) and (4x+3) using FOIL method $= (4x+3)^{-2}(-8+10[ (5x)(4x) +(5x)(3) +(1)(4x) +(1)(3) ])$ $= (4x+3)^{-2}(-8+10[ 20x^{2} +15x + 4x + 3 ])$ $= (4x+3)^{-2}(-8+10[ 20x^{2} +19x + 3 ])$ $= (4x+3)^{-2}(-8+ 200x^{2} +190x + 30)$ $= (4x+3)^{-2}( 200x^{2} +190x + 22)$ Take out common factor 2 from $( 200x^{2} +190x + 22)$ $= (4x+3)^{-2}[2( 100x^{2} +95x + 11)]$ $= \frac{2( 100x^{2} +95x + 11)}{ (4x+3)^{2}}$ $[a^{-m} = \frac{1}{a^{m}}]$
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