Without actual division,prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2$.

(N/A) Let $p(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$ and $g(x) = x^2 - 3x + 2$.
First,factorize the divisor $g(x)$:
$g(x) = x^2 - 3x + 2 = x^2 - x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)$.
For $p(x)$ to be divisible by $g(x)$,it must be divisible by both $(x - 1)$ and $(x - 2)$ by the Factor Theorem.
Check for $(x - 1)$:
$p(1) = 2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0$.
Since $p(1) = 0$,$(x - 1)$ is a factor of $p(x)$.
Check for $(x - 2)$:
$p(2) = 2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2 = 2(16) - 5(8) + 2(4) - 2 + 2 = 32 - 40 + 8 = 0$.
Since $p(2) = 0$,$(x - 2)$ is a factor of $p(x)$.
Since both $(x - 1)$ and $(x - 2)$ are factors of $p(x)$ and they are coprime,their product $(x - 1)(x - 2) = x^2 - 3x + 2$ must also be a factor of $p(x)$.
Hence,$2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2$.