Show that `(x+4),\ (x-3)a n d\ (x-7)` are factors of `x^3-6x^2-19 x+84`

Explanation

Let `f(x)=x^3-6x^2-19x+84`
If `x+4=0`, then `x=-4`
If `x-3=0`, then `x=3`
if `x-7=0`, then `x=7`
Now,
`f(-4)= (-4)^3–6(-4)^2-19(-4)+84`
`=-64-96+76+84`
`=160-160=0`
`f(-4)=0`
`f(3)=(3)^3-6(3)^2-19(3)+84`
`=27-54-57+84`
`=111-111=0`
`f(3)=0`
`f(7)=(7)^3-6(7)^2-19(7)+84`
`=343-294-133+84`
`=427–427`
`=0`
`f(7)=0`
`therefore (x+4)`, `(x–3)`,`(x-7)` are the factors of `f(x)`.