Question

What are the possible rational roots `x^5 -12x^4 +2 x^3 -3x^2 +8x-12 = 0`?

Answer 1

This quintic has no rational roots.

Explanation:

`f(x) = x^5-12x^4+2x^3-3x^2+8x-12`

By the rational root theorem, any zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-12` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-12`

Note that `f(-x) = -x^5-12x^4-2x^3-3x^2-8x-12` has all negative coefficients. Hence `f(x)` has no negative zeros.

So the only possible rational zeros are:

`1, 2, 3, 4, 6, 12`

Evaluating `f(x)` for each of these values, we find none is a zero. So `f(x)` has no rational zeros.

In common with most quintics and polynomials of higher degree, the zeros are not expressible in terms of `n`th roots or elementary functions, including trigonometric functions.

You can use numerical methods such as Durand-Kerner to find approximations:

`x_1 ~~ 11.8484`

`x_(2,3) ~~ -0.640414+-0.877123i`

`x_(4,5) ~~ 0.716229+-0.587964i`