Question

Without actual division, prove that $x^4-4x^2+12x-9$ is exactly divisible by $x^2+2x-3.$

Answer 1

Let $p\left(x\right)=x^4-4x^2+12x-9$ and $g\left(x\right)=x^2+2x-3$.

$g\left(x\right)=(x+3)(x-1)$

Hence, $(x+3)$ and $(x-1)$ are factors of $g\left(x\right)$.

In order to prove that $p\left(x\right)$ is exactly divisible by $g\left(x\right)$, it is sufficient to prove that $p\left(x\right)$ is exactly divisible by $(x+3)$ and $(x-1)$.

$\therefore$ Let us show that $(x+3)$ and $(x-1)$ are factors of $p\left(x\right)$.

Now,

$p\left(x\right)=x^4-4x^2+12x-9$

$p(-3)=(-3{)}^4-4(-3{)}^2+12(-3)-9$

$=81-36-36-9$

$=81-81$

$=0$

$\therefore p(-3)=0$

And,

$p\left(1\right)=(1{)}^4-4(1{)}^2+12\left(1\right)-9$

$=1-4+12-9$

$=13-13$

$=0$

$\therefore p\left(1\right)=0$

Now by factor theorem we can say that, $(x+3)$ and $(x-1)$ are factors of $p\left(x\right)\Rightarrow g\left(x\right)=(x+3)(x-1)$ is also factor of $p\left(x\right)$. Hence, $p\left(x\right)$ is exactly divisible by $g\left(x\right)$.