Answer
$x^{3}-10x^{2}+51x-260+ \displaystyle \frac{1300}{5+x}$
Work Step by Step
rewrite the polinomials in descending order of powers.
$(x^{4}-5x^{3}+x^{2}-5x)\div(x+5)$
Dividing with $x-c\qquad...\qquad c=-5.$
Don't forget to place 0's for the missing terms.
$\begin{array}{rrrrrrrr}
\underline{-5}| &1 & -5 & 1 & -5 &0 & \\
& & -5 & 50 &-255 &1300 & \\
& -- & -- &-- &-- &--& \\
&1 & -10 & 51 &-260 &1300 & \end{array}$
Quotient = $x^{3}-10x^{2}+51x-260$
Remainder = $1300$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{x^{2}-5x-5x^{3}+x^{4}}{5+x}$ = $x^{3}-10x^{2}+51x-260+ \displaystyle \frac{1300}{5+x}$
