Question

Determine the zeros of f(x) = x3 – 3x2 – 16x + 48

Answer 1

x3 – 3x2 – 16x + 48
x(x^2-3x-16)+48
resolves formule in x^2
x=0 and x=\[\sqrt{73}\] /2

Answer 2

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

Answer 3

zeros is: 4 or 3 or -4

Answer 4

understand?

Answer 5

yes thank you!<33