Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.2 Synthetic Division - 3.2 Exercises - Page 325: 13

Answer

$x^{4}+x^{3}+2x-1+\displaystyle \frac{3}{x+2}$

Work Step by Step

When dividing a polynomial $f(x)$ with $(x-k)$ We set up synthetic division $\quad \text{divisor } )\overline{\text{ dividend }}$, by placing $k$ in place of the divisor, and listing ALL coefficients of $f(x)$ (including the zeros ), starting from highest power of x. $k=-2$ $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & & & & & & \\ & & -- & -- & -- & -- & -- & -- & \\ & & & & & & & & \\ & & & & & & & & \end{array}$ We are ready. Bring down the leading coefficient, $1$ Multiply $(-2)(1)=-2$ and place it in the next free slot of the middle row. $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & -2 & & & & & \\ & & -- & -- & -- & -- & -- & -- & \\ & & 1 & & & & & & \\ & & & & & & & & \end{array}$ Add $3+(-2)=1$ and place $1$ in the bottom row. Multiply $(-2)(1)=-2$ and place it in the next free slot of the middle row. $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & -2 & -2 & & & & \\ & & -- & -- & -- & -- & -- & -- & \\ & & 1 & 1 & & & & & \\ & & & & & & & & \end{array}$ Add $2+(-2) =0$ and place $0$ in the bottom row. Multiply $(-2)(0)=0$ and place it in the next free slot of the middle row. $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & -2 & -2 & 0 & & & \\ & & -- & -- & -- & -- & -- & -- & \\ & & 1 & 1 & 0 & & & & \\ & & & & & & & & \end{array}$ Add $2+(0)=2$ and place it in the bottom row. Multiply $(-2)(2)=-4$ and place it in the next free slot of the middle row. $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & -2 & -2 & 0 & -4 & & \\ & & -- & -- & -- & -- & -- & -- & \\ & & 1 & 1 & 0 & 2 & & & \\ & & & & & & & & \end{array}$ Add $3+(-4)=-1$ and place it in the bottom row. Multiply $(-2)(-1)=2$ and place it in the next free slot of the middle row. $ \begin{array}{lllllllll} & & -- & -- & -- & -- & -- & -- & \\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1 & \\ & & & -2 & -2 & 0 & -4 & 2 & \\ & & -- & -- & -- & -- & -- & -- & \\ & & 1 & 1 & 0 & 2 & -1 & & \\ & & & & & & & & \end{array}$ Add $1+(2)=3$ and place it in the bottom row. $ \begin{array}{llllllll} & & -- & -- & -- & -- & -- & --\\ -2 & ) & 1 & 3 & 2 & 2 & 3 & 1\\ & & & -2 & -2 & 0 & -4 & 2\\ & & -- & -- & -- & -- & -- & --\\ & & 1 & 1 & 0 & 2 & -1 & \fbox{$3$}\\ & & & & & & & \end{array}$ Interpret the result: The last number of the bottom row represents the remainder, $3.$ The rest of the bottom row holds coefficients of the quotient, which has a degree of one less than $f(x):\quad q(x)=x^{4}+x^{3}+2x-1$ So, $\displaystyle \quad \frac{f(x)}{x-k}=q(x)+\frac{r}{x-k}$ $\displaystyle \frac{x^{5}+3x^{4}+2x^{3}+2x^{2}+3x+1}{x+2}=\quad x^{4}+x^{3}+2x-1+\frac{3}{x+2}\quad $
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