Find a rational zero of the polynomial function and use it to find all the zeros of the function.
-$$f ( x ) = 2 x ^ { 3 } - x ^ { 2 } - 6 x + 3$$
A) $\left\{ \frac { 1 } { 2 } , \sqrt { 3 } , - \sqrt { 3 } \right\}$
B) $\left\{ - \frac { 1 } { 2 } , \sqrt { 3 } , - \sqrt { 3 } \right\}$
C) $\{ 2 , \sqrt { 3 } , - \sqrt { 3 } \}$
D) $\{ - 2 , \sqrt { 3 } , - \sqrt { 3 } \}$

Question

Quizplus · Accepted Answer

By Rational Zero Theorem, the possible rational zeros are $\pm 1, \pm \frac { 1 } { 2 }, \pm 3, \pm \frac { 3 } { 2 }$. Testing these values, we find that $x = \frac { 1 } { 2 }$ is a zero. Using synthetic division, we get$$2 | \begin{array}{rrrr} 2 & -1 & -6 & 3 \\ \cline{2-4} 1 & 0 & -3 & 0 \\ \end{array}$$So, $f ( x ) = 2 x ^ { 3 } - x ^ { 2 } - 6 x + 3 = ( x - \frac { 1 } { 2 } ) ( 2 x ^ { 2 } - 3 )$. The other zeros are $x = \pm \sqrt { \frac { 3 } { 2 } }$.

Find a Rational Zero of the Polynomial Function and Use f(x)=2x3−x2−6x+3f ( x ) = 2 x ^ { 3 } - x ^ { 2 } - 6 x + 3f(x)=2x3−x2−6x+3

Find a Rational Zero of the Polynomial Function and Use $f ( x ) = 2 x ^ { 3 } - x ^ { 2 } - 6 x + 3$