# How do you use the rational roots theorem to find all possible zeros of f(x) = 6x^4 + 4x^3 - 2x^2 + 2?

Mar 31, 2016

There are no Real zeros for the given equation.

#### Explanation:

If we attempt to use the Rational Root Theorem by testing
$\textcolor{w h i t e}{\text{XXX")x=+-abs("factors of " 2)/abs("factors of } 6}$
in $f \left(x\right) = 6 {x}^{44} {x}^{3} - 2 {x}^{2} + 2$
the candidate values would be:
$\textcolor{w h i t e}{\text{XXX}} \pm \left\{1 , \frac{1}{2} , \frac{1}{3} , \frac{1}{6} , \cancel{\frac{1}{2}} , \cancel{\frac{2}{2}} , \frac{2}{3} , \cancel{\frac{2}{6}}\right\}$
but as indicated in the table below, none of these give a zero result: Therefore there are no rational zeros for this expression.

In fact, if we consider the graph for this expression, we can see that it has no Real zeros.
graph{6x^4+4x^3-2x+2 [-8.036, 9.754, -0.84, 8.05]}

$6 {x}^{4} + 4 {x}^{3} - 2 {x}^{2} + 2$ should have $4$ imaginary zeroes, but they can not be determined using the Rational Factor Theorem.