Given the Polynomial Function F(x), Find the Rational Zeros, Then $$f ( x ) = \frac { 1 } { 2 } x ^ { 3 } - \frac { 2 } { 3 } x ^ { 2 } - \frac { 1 } { 12 } x + \frac { 1 } { 12 }$$

question 47

Multiple Choice

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

• A) \(\frac { 1 } { 3 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)\)
• B) \(\frac { 1 } { 2 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( \frac { 1 } { 3 } \right) \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)\)
• C) \(\frac { 1 } { 2 } , \frac { 1 + \sqrt { 2 } } { 2 } , \frac { 1 - \sqrt { 2 } } { 2 } ; f ( x ) = \left( x - \frac { 1 } { 2 } \right) \left( x - \frac { 1 + \sqrt { 2 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 2 } } { 2 } \right)\)
• D) \(\frac { 1 } { 3 } , \frac { 1 + \sqrt { 3 } } { 2 } , \frac { 1 - \sqrt { 3 } } { 2 } ; f ( x ) = \left( \frac { 1 } { 2 } \right) \left( x - \frac { 1 } { 3 } \right) \left( x - \frac { 1 + \sqrt { 3 } } { 2 } \right) \left( x - \frac { 1 - \sqrt { 3 } } { 2 } \right)\)

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