Question

1. Determine the zeros of f(x) = x3 – 3x2 – 16x + 48? Somebody explain it please?:/

Answer 1

$x^{2} - 3x^{2} - 16x + 48 = 0$

$x^{2} (x - 3) - 16(x - 3) = 0$

$(x - 3) ( x^{2} - 16) = 0$

$(x - 3) (x - 4) (x + 4) = 0$

$x = -4, 3, 4$

The zeros are: -4, 3, 4