# How do you find all the zeros of f(x)=x^5+3x^3-x+6?

Jun 22, 2016

Find there are no rational zeros.

Use Durand-Kerner or similar to find approximations.

#### Explanation:

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

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 $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 6$

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

In common with most quintic polynomials, the zeros are not expressible in terms of $n$th roots, so you are left with numerical methods such as Durand-Kerner to help find approximations:

$x \approx - 1.17826$

$x \approx - 0.202554 \pm 1.89313 i$

$x \approx 0.791684 \pm 0.882051 i$

See https://socratic.org/s/avxUUEiJ for another example.

Here's a sample C++ program that implements the Durand-Kerner algorithm for this example: 