Question

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

Answer 1

To prove :

$x^4+2x^3-2x^2+2x-3$ is exactly divisible by $x^2+2x-3$

Proof :

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

And, $g\left(x\right)=x^2+2x-3$

Then, $g\left(x\right)=x^2+2x-3$

$=x^2+3x-x-3=x(x+3)-(x+3)=(x+3)(x-1)$

Now, we check if $g\left(x\right)$ is a factor of $p\left(x\right)$ by using factor theorem.

$\therefore$ $(x+3)$ and $(x-1)$ divides $p\left(x\right)$ if$p(-3)$ and $p\left(1\right)=0$

So,

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

$=81-54-18-6-3=0$

and,

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

$=1+2-2+2-3=0$

Hence, $p\left(x\right)$ is divisible by $(x+3)$ and $(x-1)$

$\Rightarrow p\left(x\right)$ is divisible by $(x+3)(x-1)$

$\Rightarrow p\left(x\right)$ is divisible by $g\left(x\right)$

Hence proved.