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

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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?

The Real zeros of such a quintic polynomial may be integers, fifth roots of integers (if

##### 1 Answer

Quite varied...

#### Explanation:

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

If

For example, let

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

For example, with

#(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

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