# How do you find all zeros of the function f(x) = 3x^5 - x^2 + 2x + 18?

Aug 6, 2016

Use a numerical method to find approximations for the zeros:

${x}_{1} \approx - 1.35047$

${x}_{2 , 3} \approx - 0.506174 \pm 1.35526 i$

${x}_{4 , 5} \approx 1.18141 \pm 0.852685 i$

#### Explanation:

$f \left(x\right) = 3 {x}^{5} - {x}^{2} + 2 x + 18$

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 $18$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm 9 , \pm 18$

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

In common with most quintics and higher order polynomials, the zeros are not expressible in terms of $n$th roots or elementary functions, including trigonometric ones.

About the best you can do is use a numerical method like Durand-Kerner to find approximations:

${x}_{1} \approx - 1.35047$

${x}_{2 , 3} \approx - 0.506174 \pm 1.35526 i$

${x}_{4 , 5} \approx 1.18141 \pm 0.852685 i$

See https://socratic.org/s/awNxzXZ9 for a description of the method and another example quintic approximated using this method.