# 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?

Jan 5, 2017

Quite varied...

#### Explanation:

$f \left(x\right) = {x}^{5} + p x + q$

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

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

For example, let ${x}^{2} + a x + b$ be any monic quadratic with integer coefficients.

Then note that:

$\left({x}^{2} + a x + b\right) \left({x}^{3} - a {x}^{2} + \left({a}^{2} - b\right) x + a \left(2 b - {a}^{2}\right)\right)$

$= {x}^{5} + \left(3 {a}^{2} b - {b}^{2} - {a}^{4}\right) x + a b \left(2 b - {a}^{2}\right)$

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:

$\left({x}^{2} - 3 x + 2\right) \left({x}^{3} + 3 {x}^{2} + 7 x + 15\right) = {x}^{5} - 31 x + 30$

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:

${x}^{5} + 4 x + 2$

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.