## anonymous 5 years ago Determine the zeros of f(x) = x3 – 3x2 – 16x + 48

1. anonymous

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

2. anonymous

$x^3-3 x^2-16 x+48=(x-4) (x-3) (x+4)$

3. anonymous

zeros is: 4 or 3 or -4

4. anonymous

understand?

5. anonymous

yes thank you!<33