Answer
$4x^{2}+x+4+ \displaystyle \frac{3}{x-1}$
Work Step by Step
Dividing with $x-c\qquad...\qquad c=1.$
$\begin{array}{rrrrrr}
\underline{1}| &4 &-3&3 &-1 &\\
& &4 &1 &4 &\\
& --&--&--&--&\\
&4 &1 &4 &3 &\end{array}$
Quotient = $4x^{2}+x+4$
Remainder = $3$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{4x^{3}-3x^{2}+3x-1}{x-1}$ = $4x^{2}+x+4+ \displaystyle \frac{3}{x-1}$
