How do you use the rational roots theorem to find all possible zeros of 3x^5-7x^2+x+6?

May 24, 2016

This quintic has no rational zeros.

Explanation:

$f \left(x\right) = 3 {x}^{5} - 7 {x}^{2} + x + 6$

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 $6$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

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

Evaluating $f \left(x\right)$ for each of these, we find none work.

So this quintic has no rational zeros.

In fact it has one Real zero at $x \approx - 0.873$ and two pairs of Complex zeros.

These zeros are not expressible in terms of $n$th roots, so about the best you can do is find numerical approximations using Newton's method or the Durand-Kerner method.