On dividing $x^{3}-3 x^{2}+x+2$ by a polynomial $g(x)$,the quotient and remainder were $x-2$ and $-2 x+4$,respectively. Find $g(x)$.

(N/A) Given:
Dividend $p(x) = x^{3}-3 x^{2}+x+2$
Quotient $q(x) = x-2$
Remainder $r(x) = -2 x+4$
Divisor $g(x) = ?$
Using the division algorithm for polynomials:
$p(x) = g(x) \cdot q(x) + r(x)$
$x^{3}-3 x^{2}+x+2 = g(x) \cdot (x-2) + (-2 x+4)$
Rearranging the terms to isolate $g(x) \cdot (x-2)$:
$g(x) \cdot (x-2) = (x^{3}-3 x^{2}+x+2) - (-2 x+4)$
$g(x) \cdot (x-2) = x^{3}-3 x^{2}+x+2+2 x-4$
$g(x) \cdot (x-2) = x^{3}-3 x^{2}+3 x-2$
Now,divide $(x^{3}-3 x^{2}+3 x-2)$ by $(x-2)$ to find $g(x)$:
By performing polynomial long division:
$(x^{3}-3 x^{2}+3 x-2) \div (x-2) = x^{2}-x+1$
Therefore,$g(x) = x^{2}-x+1$.