Question

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

Answer 1

Find `f(x)` has no rational zeros.

We can find numerical approximations:

`x_1 ~~ 1.6477`

`x_(2,3) ~~ 0.151952+-1.03723i`

`x_(4,5) ~~ -0.975801+-1.12111i`

Explanation:

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`f(x) = x^5-3x^2-4`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `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:

`+-1, +-2, +-4`

None of these work, so `f(x)` 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 ~~ 1.6477`

`x_(2,3) ~~ 0.151952+-1.03723i`

`x_(4,5) ~~ -0.975801+-1.12111i`

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