Question

Show that $(x-2),(x+3)$ and $(x-4)$ are factors of $x^3-3x^2-10x+24$.

Answer 1

Given, $(x-2)$ , $(x+3)$ and $(x-4)$ are factors of polynomial $x^3-3x^2-10x+24$.

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

If $(x-2)$ is a factor, then $x-2=0⟹x=2$.

Replace $x$ by $2$, we get,

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

$f\left(2\right)=8-12-20+24$

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

The value of $f\left(2\right)$ is zero.

Then $(x-2)$ is the factor of the polynomial $x^3-3x^2-10x+24$.

If $(x+3)$ is factor, then $x+3=0⟹x=-3$.

Replace $x$ by $-3$, we get,

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

$f(-3)=-27-27+30+24$

$f(-3)=0$.

The value of $f(-3)$ is zero.

Then $(x+3)$ is the factor of the polynomial $x^3-3x^2-10x+24$.

If $(x-4)$ is factor, then $x-4=0⟹x=4$.

Replace $x$ by $4$, we get,

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

$f\left(4\right)=64-48-40+24$

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

The value of $f\left(4\right)$ is zero.

Then $(x+4)$ is the factor of the polynomial $x^3-3x^2-10x+24$.

Therefore, hence showed that $(x-2)$ , $(x+3)$ and $(x-4)$ are factors of polynomial $x^3-3x^2-10x+24$.