Answer
$x^{4}+7x^{3}+21x^{2}+60x+182+ \displaystyle \frac{549}{x-3}$
Work Step by Step
Dividing with $x-c\qquad...\qquad c=3.$
Don't forget to place 0's for the missing terms.
$\begin{array}{rrrrrrrr}
\underline{3}| &1 & 4 & 0 & -3 &2 &3 \\
& & 3 &21 &63 &180 &546 \\
& --& --&--&-- &--&--\\
&1 & 7 &21 &60 &182 & 549 \end{array}$
Quotient = $x^{4}+7x^{3}+21x^{2}+60x+182$
Remainder = $549$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{x^{5}+4x^{4}-3x^{2}+2x+3}{x-3}$ = $x^{4}+7x^{3}+21x^{2}+60x+182+ \displaystyle \frac{549}{x-3}$
