Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.4 - Division of Polynomials - Exercise Set - Page 444: 27

Answer

$2x^{2}+4x+5+\displaystyle \frac{9}{2x+3}$

Work Step by Step

$\begin{array}{ccccccccccc} & &2x^{2} & +4x& +5 & \\ & &--&-- &--& \\ 2x-4&) & 4x^{3} & & -6x& -11 & \color{blue}{\leftarrow \small{\text{no }x^{2}... } } \\ & & 4x^{3} & -8x^{2}& & & \color{red}{\leftarrow \small{2x^{2}(2x-43) } } \\ & &--&-- & & & \color{red}{ \small{subtract}} \\ & & & 8x^{2}& -6x& -11 & & \\ & & & 8x^{2} & -16x& & \color{red}{\leftarrow \small{4x(2x-4) } }\\ & & &--&-- & &\color{red}{ \small{subtract}} \\ & & & & 10x & -11 & \\ & & & & 10x & -20 & \color{red}{\leftarrow \small{5(2x-4) } }\\ & & & &-- & -- &\color{red}{ \small{subtract}} \\ & & & & & 9 & \end{array}$ Quotient = $2x^{2}+4x+5$ Remainder = $ 9$. $\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$ $\displaystyle \frac{ 4x^{3}-6x-11}{2x-4}$ = $2x^{2}+4x+5+\displaystyle \frac{9}{2x+3}$
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