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

Aug 5, 2016

This quartic has no rational zeros. It has $4$ non-Real Complex zeros:

${x}_{1 , 2} \approx 0.56707 \pm 0.43127 i$

${x}_{3 , 4} \approx - 0.73373 \pm 0.34406 i$

#### Explanation:

$f \left(x\right) = 6 {x}^{4} + 2 {x}^{3} - 3 {x}^{2} + 2$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $6$ of the leading term $6$.

So the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm 2$

Trying any of these, we find $f \left(x\right) > 0$, so $f \left(x\right)$ has no rational zeros.

In fact, this quartic has only Complex zeros, which it is possible, but horribly messy to find algebraically. We can use numerical methods such as Newton's method or Durand-Kerner to find approximations:

${x}_{1 , 2} \approx 0.56707 \pm 0.43127 i$

${x}_{3 , 4} \approx - 0.73373 \pm 0.34406 i$