Question

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

Answer 1

$x^2-3x+2$

$\Rightarrow x^2-x-2x+2$

$\Rightarrow x(x-1)-2(x-1)$

$\Rightarrow (x-1)(x-2)$

$\Rightarrow (x-1)(x-2)$ are the factors of $x^2-3x+2$

If we prove $(x-1)and(x-2)$are the factors of $2x^4+2x^2-x+2$, then $2x^4+2x^2-x+2$

is divisible by $x^2-3x+2$

Now,Let $F\left(x\right)=2x^4+2x^2-x+2$

$\Rightarrow F\left(1\right)=2(1{)}^4+2(1{)}^2-1+2$

$\Rightarrow F\left(1\right)=5\ne 0$

So,$2x^4+2x^2-x+2$ is not divisible by $x^2-3x+2$