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 445: 36

Answer

$6x^2+3x-1+\dfrac{-3x+1}{3x^2+1}$

Work Step by Step

$ \begin{array}{llllllllll} & & 6x^{2} & +3x & -1 & & & \color{red}{ \small{Quotient}} & & \\ & & -- & -- & -- & -- & -- & -- & & \\ 3x^{2}+1 & ) & 18x^{4} & +9x^{3} & +3x^{2} & +0 & +0 & \color{blue}{\leftarrow \small{ no\ a_{1}x^{1}+a_{0}... } } & & \\ & & 18x^{4} & & +6x^{2} & & & \color{red}{\leftarrow \small{ 6x^{2}(3x^{2}+1) } } & & \\ & & -- & -- & -- & & & \color{red}{\leftarrow \small{ subtract } } & & \\ & & & 9x^{3} & -3x^{2} & +0 & +0 & & & \\ & & & 3x^{3} & & +3x & & \color{red}{\leftarrow \small{3x(3x^{2}+1) } } & & \\ & & & -- & -- & -- & & \color{red}{\leftarrow \small{ subtract }} & & \\ & & & & -3x^{2} & -3x & & & & \\ & & & & -3x^{2} & & -1 & \color{red}{\leftarrow \small{ -1(3x^{2}+1) } } & & \\ & & & & -- & -- & -- & \color{red}{\leftarrow \small{ subtract } } & & \\ & & & & & & & & & \\ & & & & & -3x & +1 & \color{red}{\leftarrow \small{ Remainder } } & & \end{array} $ $(15x^{4}+3x^{3}+4x^{2}+4)\div(3x^{2}-1)=$ $=6x^2+3x-1+\dfrac{-3x+1}{3x^2+1}$
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