Answer
$x^{3}-12x^{2}+73x-444+ \displaystyle \frac{2664}{6+x}$
Work Step by Step
Rewrite the polynomials in descending order of powers.
$(x^{4}-6x^{3}+x^{2}-6x)\div(x+6)$
Dividing with $x-c\qquad...\qquad c=-6.$
Don't forget to place 0's for the missing terms.
$\begin{array}{rrrrrrrr}
\underline{-6}| &1 & -6 & 1 & -6 &0 & \\
& & -6 & 72 &-438 &2664 & \\
& -- & -- &-- &-- &-- & \\
&1 & -12 & 73 &-444 &2664 & \end{array}$
Quotient = $x^{3}-12x^{2}+73x-444$
Remainder = $2664$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{x^{2}-6x-6x^{3}+x^{4}}{6+x}$ = $x^{3}-12x^{2}+73x-444+ \displaystyle \frac{2664}{6+x}$
