Use the rational zeros theorem to determine the potential rational zeros of the polynomial function. Do not find the zeros.
-$$f ( x ) = 6 x ^ { 4 } + 4 x ^ { 3 } - 3 x ^ { 2 } + 2$$
A) $\pm \frac { 1 } { 6 } , \pm \frac { 1 } { 3 } , \pm \frac { 1 } { 2 } , \pm 1 , \pm 2$
B) $\pm \frac { 1 } { 6 } , \pm \frac { 1 } { 3 } , \pm \frac { 1 } { 2 } , \pm \frac { 2 } { 3 } , \pm 1 , \pm 2$
C) $\pm \frac { 1 } { 2 } , \pm \frac { 3 } { 2 } , \pm 1 , \pm 2 , \pm 3 , \pm 6$
D) $\pm \frac { 1 } { 6 } , \pm \frac { 1 } { 3 } , \pm \frac { 1 } { 2 } , \pm \frac { 2 } { 3 } , \pm 1 , \pm 2 , \pm 3$

Question

Quizplus · Accepted Answer

The Rational Zeros Theorem states that any rational zero of a polynomial function $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ will have the form $\pm \frac{p}{q}$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$. For $f(x) = 6x^4 + 4x^3 - 3x^2 + 2$, the constant term $a_0 = 2$ and the leading coefficient $a_n = 6$. The factors of 2 are $\pm 1, \pm 2$ and the factors of 6 are $\pm 1, \pm 2, \pm 3, \pm 6$. Therefore, the potential rational zeros are $\pm \frac{1}{6}, \pm \frac{1}{3}, \pm \frac{1}{2}, \pm \frac{2}{3}, \pm 1, \pm 2$.

Use the Rational Zeros Theorem to Determine the Potential Rational f(x)=6x4+4x3−3x2+2f ( x ) = 6 x ^ { 4 } + 4 x ^ { 3 } - 3 x ^ { 2 } + 2f(x)=6x4+4x3−3x2+2

Use the Rational Zeros Theorem to Determine the Potential Rational $f ( x ) = 6 x ^ { 4 } + 4 x ^ { 3 } - 3 x ^ { 2 } + 2$