Question

What are all the possible rational zeros for `f(x)=4x^5+4x^2-9` and how do you find all zeros?

Answer 1

This quintic has no rational zeros. It has one irrational Real zero and `4` non-Real complex ones.

Explanation:

`f(x) = 4x^5+4x^2-9`

By the rational root 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 `-9` and `q` a divisor of the coefficient `4` of the leading therm.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-9/4, +-3, +-9/2, +-9`

We find:

`f(1) = 4+4-9 = -1 < 0`

`f(3/2) = 4(243/32)+4(9/4)-9 = 243/8 > 0`

So `f(x)` has an irrational zero in `(1, 3/2)`

Looking at the derivative of `f(x)` we find:

`f'(x) = 20x^4+8x = 4x(5x^3+2)`

So `f(x)` has turning points at `x=0` and `x=-root(3)(2/5)`

`f(0) = -9 < 0`

`f(-root(3)(2/5)) = 4(root(3)(2/5))^2(2/5+1)-9 = 28/5(root(3)(2/5))^2-9 < 6-9 < 0`

So the only Real zero of `f(x)` is the irrational one in `(1, 3/2)`

Typically for a quintic, it is not possible to express the zeros in terms of radicals or elementary functions (including trigonometric, logarithmic or exponential functions).

The other `4` Complex zeros will occur as two pairs of complex conjugates.

It is possible to find numerical approximations using algorithms such as Durand-Kerner.