Question

Determine the zeros of f(x) = x^3 – 3x^2 – 16x + 48

Answer 1

If we factor by grouping, we get the following\[x^3-3x^2-16x+48\]\[x^2(x-3)-16(x-3)\]\[(x-3)(x^2-16)\]\[(x-3)(x+4)(x-4)\]So we have zeroes of \(3, 4, -4\)

Answer 2

f(x) = x3 – 3x2 – 16x + 48
x^2(x-3)-16(x-3)
(x-3)(x^2-16)
(x-3)(x-4)(x+4)
for f(x) =0
f(x) =(x-3)(x-4)(x+4) =0
x =3 , 4 , -4