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

Aug 8, 2016

This quartic has no rational zeros, but we can find approximations:

${x}_{1} \approx 0.694795$

${x}_{2} \approx - 0.800431$

${x}_{3 , 4} \approx - 0.113849 \pm 1.08894 i$

#### Explanation:

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

By the rational root theorem and rational zeros of $P \left(x\right)$ 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 $3$ of the leading term.

That means that the only possible rational zeros are:

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

None of these works, so $P \left(x\right)$ has no rational zeros.

As a quartic, it is possible to solve $P \left(x\right)$ algebraically, but typically for a quartic, it gets very messy. It is much simpler to find approximations to the zeros using a numerical method such as Durand-Kerner.

Here's a C++ program for this quartic... Hence we can find approximations:

${x}_{1} \approx 0.694795$

${x}_{2} \approx - 0.800431$

${x}_{3 , 4} \approx - 0.113849 \pm 1.08894 i$