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Is it possible to find a 5th degree polynomial with no rational roots, and at least one irrational root?

Edit: the polynomial must only have rational coefficients

In fact we can even achieve this with a polynomial with rational coefficients: [Edit: After I wrote this answer, OP added the rationality of the coefficients as a condition.]

The Rational Root Theorem implies that $$x^5 - 2$$ has no rational roots, but since its degree is odd, it has at least one real---and hence irrational---root. (In fact, can replace $2$ here with any rational number that is not a perfect $5$th power of a rational number.)

That is pretty easy, if there are not other propertys required. Just consider a polynomial like:

$(x-sqrt2)^5$

If $alpha$ is irrational, then the polynomial $(x-alpha)^5$ has degree $5$, and the only (repeated) root is irrational.

$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 $$

Five real irrational roots,
$$ 2 cos left( frac2k
pi11 right) $$

If you prefer the roots to be $cos( 2k pi /11),$ adjust the quintic to

$$ 32x^5 + 16x^4 - 32 x^3 - 12 x^2 + 6 x + 1 $$

gp-pari:

? x = cos( 2 * Pi / 11)
%1 = 0.8412535328311811688618116489
? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1
%2 = -1.009741959 E-28
? 
? x = cos( 4 * Pi / 11)
%3 = 0.4154150130018864255292741493
? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1
%4 = -2.019483917 E-28
? 
? x = cos( 6 * Pi / 11)
%5 = -0.1423148382732851404437926686
? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1
%6 = -7.57306469 E-29
? 
? x = cos( 8 * Pi / 11)
%7 = -0.6548607339452850640569250725
? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1
%8 = 0.E-28
? 

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Well, $sqrt[5]m$ is irrational if $m$ is an integer that isn't a perfect power of $5$.

And $sqrt[5]m$ is a solution to $x^5 -m =0$.

And as $x^5 -m =0implies x^5 = mimplies x =sqrt [5]m$ so that is the only solution.