Question

If `f(x) = x^5+px+q` (Bring Jerrard normal form) with `p, q` integers, then what are the possible natures of the zeros?

The Real zeros of such a quintic polynomial may be integers, fifth roots of integers (if `p = 0`) or not expressible in terms of radicals. Are there any other possibilities?

Answer 1

Quite varied...

Explanation:

`f(x) = x^5+px+q`

Bring-Jerrard normal form is - perhaps surprisingly - not particularly restrictive.

If `q != 0`, then by Descartes' Rule of Signs we can deduce that `f(x)` has at most `3` real zeros. Apart from this maximum number of Real zeros, there is quite a variety of possibilities.

For example, let `x^2+ax+b` be any monic quadratic with integer coefficients.

Then note that:

`(x^2+ax+b)(x^3-ax^2+(a^2-b)x+a(2b-a^2))`

`= x^5+(3a^2b-b^2-a^4)x + ab(2b-a^2)`

which is in Bring Jerrard normal form.

The cubic will have at least one Real zero, so we can choose `a, b` to give our quintic up to `3` Real zeros.

For example, with `a=-3` and `b=2` we find:

`(x^2-3x+2)(x^3+3x^2+7x+15) = x^5-31x+30`

The cubic can be solved by Cardano's method to find its three zeros (one Real, two Complex) in terms of ordinary radicals.

Alternatively, if we choose `b > 0` much larger than `a` then we can arrive at a quadratic with non-Real zeros and a cubic with `3` Real zeros. See https://socratic.org/s/aBai7g2H

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By way of contrast:

`x^5+4x+2`

has one irrational negative Real zero and four complex zeros, occurring in complex conjugate pairs. None of these zeros is expressible in terms of elementary functions, including `n`th roots, trigonometric, exponential or logarithmic functions.