Answer
The quotient is $3x^2+11x+32$
and the remainder is $99$.
Work Step by Step
The given expression is:-
$(3x^3+2x^2-x+3)\div (x-3)$
The divisor is $x-3$, so the value of $c=3$.
and on the right side the coefficients of dividend in descending powers of $x$.
Perform the synthetic division to obtain:
$\begin{matrix}
&-- &-- &--&--& \\
3) &3&2&-1&3& & \\
& &9 &33 &96 && \\
& -- & -- & --& -- && \\
& 3 & 11& 32 &99 & \\
\end{matrix}$
The divisor is $x-3$
The dividend is $3x^3+2x^2-x+3$
The Quotient is $3x^2+11x+32$
The remainder is $99$.
Check:-
$=(\text{Divisor)(Quotient)+Remainder}$
$=(x-3)(3x^2+11x+32)+99$
$=3x^3+11x^2+32x-9x^2-33x-96+99$
$=3x^3+2x^2-x+3$
Hence, the quotient is $3x^2+11x+32$ and the remainder is $99$.
