Answer
$x^{6}-2x^{5}+5x^{4}-10x^{3}+10x^{2}-20x+40-\displaystyle \frac{68}{x+2}$
Work Step by Step
Dividing with $x-c\qquad...\qquad c=-2.$
Don't forget to place 0's for the missing terms.
$\begin{array}{rrrrrrrr}
\underline{-2}| &1 & 0 & 1 & 0 & -10 & 0 &0 & 12 \\
& & -2 & 4 &-10 & 20 & -20 & 40 &-80 \\
& --& -- &-- &-- &-- & -- &-- & -- \\
&1 & -2 & 5 & -10 & 10 & -20 & 40 &-68 \end{array}$
Quotient = $x^{6}-2x^{5}+5x^{4}-10x^{3}+10x^{2}-20x+40$
Remainder = $-68$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{x^{7}+x^{5}-10x^{3}+12}{x+2}$ = $x^{6}-2x^{5}+5x^{4}-10x^{3}+10x^{2}-20x+40-\displaystyle \frac{68}{x+2}$
