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

Explanation

Let `f(x)=x^3-3x^2-10x+24`
By Factor Theorem:
If `p(x)` is a polynomial and a is any real number, then `g(x)=(x– a)` is a factor of `p(x)`, if `p(a)=0`
For checking `(x-2)` to be a factor, we will find `f(2)`
`⇒f(2)=(2)^3-3(2)^2-10(2)+24`
`⇒f(2)=8-12-20+24`
`⇒f(2)=0`
So, `(x–2)` is a factor.
For checking `(x+3)` to be a factor, we will find `f(–3)`
`⇒f(–3)=(–3)^3–3(–3)^2–10(–3)+24`
`⇒f(-3)=-27-27+30+24`
`⇒f(–3)=0`
So, `(x+3)` is a factor.
For checking `(x – 4)` to be a factor, we will find `f(4)`
`⇒f(4)=(4)^3–3(4)^2–10(4)+24`
`⇒f(4)=64–48–40+24`
`⇒f(4)=0`
So, `(x–4)` is a factor.
`∴ (x–2)`, `(x+3)` and `(x–4)` are factors of `x^3 -3x^2-10x+24`