## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 6 - Section 6.4 - Division of Polynomials - Exercise Set - Page 445: 54

#### Answer

$x=a^{2} -3a+2$

#### Work Step by Step

Subtract $6$ from both sides and factor x out on the LHS. $x(a-3)=a^{3}-6a^{2}+11a-6\quad$... divide with $(a+2)$ $x=\displaystyle \frac{a^{3}-6a^{2}+11a-6}{a-3}\quad$... perform long division $\begin{array}{llllllll} & & a^{2} & -3a & +2 & & & \color{red}{ \small{Quotient}} \\ & & -- & -- & -- & -- & & \\ a-3 & ) & a^{3} & -6a^{2} & +11a & -6 & & \\ & & a^{3} & -3a^{2} & & & & \color{red}{\leftarrow \small{ a^{2}(a-3) } } \\ & & -- & -- & & & & \color{red}{\leftarrow \small{ subtract } } \\ & & & -3a^{2} & +11a & -6 & & \\ & & & -3a^{2} & +9a & & & \color{red}{\leftarrow \small{-3a(a-3) } }\\ & & & -- & -- & & & \color{red}{\leftarrow \small{ subtract }} \\ & & & & 2a & -6 & & \\ & & & & 2a & -6 & & \color{red}{\leftarrow \small{2(a-3) } }\\ & & & & -- & -- & & \color{red}{\leftarrow \small{ subtract }} \\ & & & & & 0 & & \color{red}{\leftarrow \small{ Remainder } } \end{array}$ $x=a^{2} -3a+2$

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