Given the polynomial function f(x), find the rational zeros, then the other zeros (that is, solve the equation f(x) = 0), and
factor f(x) into linear factors.
-

f ( x ) = \frac { 1 } { 2 } x ^ { 3 } - \frac { 2 } { 3 } x ^ { 2 } - \frac { 1 } { 12 } x + \frac { 1 } { 12 }

\frac { 1 } { 3 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)

\frac { 1 } { 2 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( \frac { 1 } { 3 } \right) \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)

\frac { 1 } { 2 } , \frac { 1 + \sqrt { 2 } } { 2 } , \frac { 1 - \sqrt { 2 } } { 2 } ; f ( x ) = \left( x - \frac { 1 } { 2 } \right) \left( x - \frac { 1 + \sqrt { 2 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 2 } } { 2 } \right)

\frac { 1 } { 3 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( \frac { 1 } { 2 } \right) \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)

Given That the Polynomial Function Has the Given Zero, Find f(x)=x3−2x2−11x+52;−4f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 11 x + 52 ; - 4f(x)=x3−2x2−11x+52;−4

Multiple Choice

Correct Answer:VerifiedUnlock this answer nowGet Access to more Verified Answers free of chargeAccess For Free

Given That the Polynomial Function Has the Given Zero, Find $f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 11 x + 52 ; - 4$

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge