OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

zeros is: 4 or 3 or -4

5 years ago
OpenStudy (anonymous):

understand?

5 years ago
OpenStudy (anonymous):

yes thank you!<33

5 years ago
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