Question

How do you find all the zeros of `f(x) = 9x^4 – 17x^3 + 2x^2 – 3x + 33`?

Answer 1

Check that it has no rational zeros, then resort to a numerical method to find approximations.

Explanation:

`f(x) = 9x^4-17x^3+2x^2-3x+33`

By the rational root theorem, any rational zeros of `f(x)` can be expressed in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `33` and `q` a divisor of the coefficient `9` of the leading term.

So the only possible rational zeros are:

`+-1/9`, `+-1/3`, `+-1`, `+-11/9`, `+-3`, `+-11/3`, `+-11`, `+-33`

In addition, notice that if you invert the signs of the coefficients on the terms of odd degree, then all of the resulting coefficients are positive. Hence `f(x)` has no negative zeros.

So the only possible rational zeros are:

`1/9, 1/3, 1, 11/9, 3, 11/3, 11, 33`

We find that `f(x) > 0` for all of these values of `x`. So `f(x)` has no rational zeros.

It is possible to solve such a quartic algebraically, but it gets very messy and is not necessarily very useful.

You can use a numerical method such as Durand-Kerner to find good approximations to the two pairs of Complex zeros that it does have.

See https://socratic.org/s/av2PtWLY for an example finding approximations for the zeros of a quintic.

In our example, we find approximations:

`x = 1.57163 +-0.656656i`

`x = -0.627188 +-0.932989i`

An example C++ program that performs the calculations would be: