Question

Prove that $(x-2),(x+3)$ and $(x-4)$ are factors of the polynomial: $p\left(x\right)=x^3-3x^2-10x+24$.

Answer 1

We have,

$f\left(x\right)=x^3-3x^2-3x^2-10x+24$

Now,

$x-2,x+3,x-4$

If, $x-2=0$

$x=2$ Put and we get,

$f\left(2\right)=2^3-3{\left(2\right)}^2-10\left(2\right)+24$

$=8-12-20+24$

$=32-32$

$f\left(2\right)=0$

If, $x+3=0$

$x=-3$

So put

$f(-3)={(-3)}^3-3{(-3)}^2-10(-3)+24$

$=-27-27+30+24$

$=-54+54$

$=0$

If $x-4=0$

$x=4\text{put}$

$f\left(x\right)=x^3-3x^2-10x+24$

$f\left(4\right)={\left(4\right)}^3-3{\left(4\right)}^2-10\times 4+24$

$=64-48-40+24$

$=88-88$

$f\left(4\right)=0$

These are factors of polynomial

Hence, proved.