Without actual division, prove that

Let $f(x)=2 x^{4}-5 x^{3}+2 x^{2}-x+2$ and $g(x)=x^{2}-3 x+2$

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

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

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

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

Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).

Putting x = 1 in f(x), we get

$f(1)=2 \times 1^{4}-5 \times 1^{3}+2 \times 1^{2}-1+2=2-5+2-1+2=0$

By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).

Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).

Hence, $f(x)=2 x^{4}-5 x^{3}+2 x^{2}-x+2$ is exactly divisible by $(x-1)(x-2)=x^{2}-3 x+2$