Is a polynomial of degree 3 with irrational roots possible?

Question

It is easy to give an example of a polynomial of degree 3 with integer coefficients having:

(a) three distinct rational roots,

(b) one rational root and two irrational roots.

But for a while I am trying to construct one that all its roots are irrational but I can't. It seems that it is not possible at all?

Also, can a polynomial with integer coefficients of degree 3 have two rational roots and one irrational root?

Answer 1

Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.

Answer 2

To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.

Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+\sqrt{b}$ is $a-\sqrt{b}$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.

For the case with all 3 irrational roots, see here.

Answer 3

Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)\cdot q(x)$$

then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)

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However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.

Answer 4

By Vieta the sum of the roots must be rational, hence this excludes a single irrational.

All other cases are possible.

$$\begin{align}0&:x(x^2-1)=0, \\2&:x(x^2-2)=0, \\3&:8x^3-6x-1=0.\end{align}$$

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The last one was built from

$$\cos3\theta=4\cos^3\theta-3\cos\theta=\cos\frac\pi3$$

so that the roots are

$$\cos\frac\pi9, \cos\frac{7\pi}9, \cos\frac{13\pi}9.$$