Question

Verify using factor theorem, whether $2x^4-6x^3+3x^2+3x-2$ is divisible by $x^2-3x+2$ or not?

Answer 1

The divisor is not a linear polynomial, it is a quadratic polynomial.

Let us factorize the divisor :

$x^2-3x+2$

$=$ $x^2-2x-x+2$

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

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

Then $(x-1)$ and $(x-2)$ are factors of polynomial $x^2-3x+2$.

Let $p\left(x\right)=2x^4-6x^3+3x^2+3x-2$

If $(x-1)$ and $(x-2)$ both are factors of $p\left(x\right)$, then for $x=1$ and $x=2$, $p\left(x\right)$ should be $0$.

On replacing $x$ by $1$, we get

$p\left(1\right)=2(1{)}^4-6(1{)}^3+3(1{)}^2+3(1)-2$

$\Rightarrow p\left(1\right)=2-6+3+3-2=0$

On replacing $x$ by $2$, we get

$p\left(2\right)=2(2{)}^4-6(2{)}^3+3(2{)}^2+3(2)-2$

$\Rightarrow p\left(2\right)=32-48+12+6-2=0$

So, $(x-1)$ and $(x-2)$ are factors of $p\left(x\right)$,

Hence, $p\left(x\right)=2x^4-6x^3+3x^2+3x-2$ is divisible by $x^2-3x+2$.