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?

1 Answer
Jan 5, 2017

Answer:

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

#color(white)()#

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.