# How do you use the rational roots theorem to find all possible zeros of f(x)=x^5-3x^2-4?

Aug 13, 2016

Find $f \left(x\right)$ has no rational zeros.

We can find numerical approximations:

${x}_{1} \approx 1.6477$

${x}_{2 , 3} \approx 0.151952 \pm 1.03723 i$

${x}_{4 , 5} \approx - 0.975801 \pm 1.12111 i$

#### Explanation:

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$f \left(x\right) = {x}^{5} - 3 {x}^{2} - 4$

By the rational roots 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 $- 4$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4$

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

In common with most quitics and polynomials of higher degree, the zeros of this one cannot be expressed in terms of $n$th roots and/or elementary functions, including trigonometric or exponential ones.

It is possible to find numerical approximations for the zeros using a method like Durand-Kerner.

For example, we can use this C++ program...

to find numerical approximations for the zeros:

${x}_{1} \approx 1.6477$

${x}_{2 , 3} \approx 0.151952 \pm 1.03723 i$

${x}_{4 , 5} \approx - 0.975801 \pm 1.12111 i$

For a little more explanation of the method see https://socratic.org/s/ax2iiWhR