Chapter 3 - Polynomial and Rational Functions - 3.2 Synthetic Division - 3.2 Exercises - Page 325: 12

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

When dividing a polynomial $f(x)$ with $(x-k)$ We set up syntehetic 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. $k=-3$ $ \begin{array}{llllllll} & & -- & -- & -- & -- & -- & \\ -3 & ) & 1 & 5 & 4 & -3 & 9 & \\ & & & & & & & \\ & & -- & -- & -- & -- & -- & \\ & & & & & & & \\ & & & & & & & \end{array}$ We are ready. Bring down the leading coefficient, $1$ Multiply $(-3)(1)=-3$ and place it in the next free slot of the middle row. $ \begin{array}{lllllll} & & -- & -- & -- & -- & --\\ -3 & ) & 1 & 5 & 4 & -3 & 9\\ & & & -3 & & & \\ & & -- & -- & -- & -- & --\\ & & 1 & & & & \\ & & & & & & \end{array}$ Add $5+(-3)=2$ and place $2$ in the bottom row. Multiply $(-3)(2)=-6$ and place it in the next free slot of the middle row. $ \begin{array}{lllllll} & & -- & -- & -- & -- & --\\ -3 & ) & 1 & 5 & 4 & -3 & 9\\ & & & -3 & -6 & & \\ & & -- & -- & -- & -- & --\\ & & 1 & 2 & & & \\ & & & & & & \end{array}$ Add $4+(-6) =-2$ and place $2$ in the bottom row. Multiply $(-3)(-2)=6$ and place it in the next free slot of the middle row. $ \begin{array}{lllllll} & & -- & -- & -- & -- & --\\ -3 & ) & 1 & 5 & 4 & -3 & 9\\ & & & -3 & -6 & 6 & \\ & & -- & -- & -- & -- & --\\ & & 1 & 2 & -2 & & \\ & & & & & & \end{array}$ Add $-3+(6)=3$ and place it in the bottom row. Multiply $(-3)(3)=-9$ and place it in the next free slot of the middle row. $ \begin{array}{lllllll} & & -- & -- & -- & -- & --\\ -3 & ) & 1 & 5 & 4 & -3 & 9\\ & & & -3 & -6 & 6 & -9\\ & & -- & -- & -- & -- & --\\ & & 1 & 2 & -2 & 3 & \\ & & & & & & \end{array}$ Add $9+(-9)=0$ and place it in the bottom row. $ \begin{array}{lllllll} & & -- & -- & -- & -- & --\\ -3 & ) & 1 & 5 & 4 & -3 & 9\\ & & & -3 & -6 & 6 & -9\\ & & -- & -- & -- & -- & --\\ & & 1 & 2 & -2 & 3 & \fbox{$0$}\\ & & & & & & \end{array}$ Interpret the result: The last number of the bottom row represents the remainder, $0.$ The rest of the bottom row holds coefficients of the quotient, which has a degree of one less than $f(x):\quad q(x)=x^{3}+2x^{2}-2x+3$ So, $\displaystyle \quad \frac{f(x)}{x-k}=q(x)+\frac{r}{x-k}$ $\displaystyle \frac{x^{4}+5x^{3}+4x^{2}-3x+9}{x+3}=\quad x^{3}+2x^{2}-2x+3+\frac{0}{x+3}\quad $ $=x^{3}+2x^{2}-2x+3$