Answer
The quotient is $x^2+x+4$ and the remainder is $12$.
Work Step by Step
The given expression is:-
$(x^3-x^2+2x+4)\div (x-2)$
The divisor is $x-2$, so the value of $c=2$.
and on the right side the coefficients of dividend in descending powers of $x$.
$\begin{matrix}
&-- &-- &--&--& \\
2) &1&-1&2&4& & \\
& &2 &2 &8 && \\
& -- & -- & --& -- && \\
& 1 & 1& 4 &12 & \\
\end{matrix}$
The divisor is $x-2$
The dividend is $x^3-x^2+2x+4$
The remainder is $12$.
The Quotient is $x^2+x+4$
Check:-
$\text{(Divisor)(Quotient)+Remainder}$
$=(x-2)(x^2+x+4)+12$
$=x^3+x^2+4x-2x^2-2x-8+12$
$=x^3-x^2+2x+4$
Hence, the quotient is $x^2+x+4$ and the remainder is $12$.
