Question

Determine the zeros of f(x) = x3 – 3x2 – 16x + 48, I already have most of it...

Answer 1

So far I know that one zero is 3, but I don't know where to go from there

Answer 2

\[x^3-3x^2-16x+48=x^2(x-3)-16(x-3)=(x-3)(x^2-16)=(x-3)(x-4)(x+4)\]

Answer 3

zeros are: -4 , 3 and 4