Question

What are all the possible rational zeros for `f(x)=x^5+2x^4+14x^3+26x^2+40x+80` and how do you find all zeros?

Answer 1

Find there are no rational zeros. Can use a numerical method to find approximations.

Explanation:

`f(x) = x^5+2x^4+14x^3+26x^2+40x+80`

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 `80` and `q` a divisor of the coefficient `1` of the leading term.

In addition note that all of the coefficients of `f(x)` are positive, so it has no positive zeros.

So the only possible rational zeros of `f(x)` are:

`-1, -2, -4, -5, -8, -10, -16, -20, -40, -80`

None of these work, so `f(x)` has no rational zeros.

Typically for a quintic or polynomial of higher degree, the zeros of `f(x)` are not expressible in terms of `n`th roots or elementary functions (including trigonometric functions).

About the best you can do is find numerical approximations using a method such as Durand-Kerner to find:

`x_1 ~~ -1.92974`

`x_(2,3) ~~ -0.135287+-3.0955i`

`x_(4,5) ~~ 0.100156+-2.07561i`

For more details on using Durand-Kerner to find such zeros of a quintic, see https://socratic.org/s/axney67c

Here's an example C++ program that implements the Durand-Kerner algorithm for our current example...