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

We have,

$x^{2}-3 x+2=x^{2}-x-2 x+2$

$=x(x-1)-2(x-1)$

$=(x-1)(x-2)$

Let $p(x)=2 x^{4}-5 x^{3}+2 x^{2}-x+2$

Now, $\quad p(1)=2(1)^{4}-5(1)^{3}+2(1)^{2}-1+2=2-5+2-1+2=0$

Therefore, $(x-1)$ divides $p(x)$

And $\quad p(2)=2(2)^{4}-5(2)^{3}+2(2)^{2}-2+2$

$=32-40+8-2+2=0$

Therefore, $(x-2)$ divides $p ( x )$.

So, $(x-1)(x-2)=x^{2}-3 x+2$ divides $2 x^{4}-5 x^{3}+2 x^{2}-x+2$