# Is a polynomial of degree 3 with irrational roots possible? [duplicate]

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?

## marked as duplicate by lulu, Winther, ℋolo, Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 4 '18 at 16:24

• @ChinnapparajR, two rationals one irrational, is my question – user231343 Oct 4 '18 at 12:44
• Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job. – lulu Oct 4 '18 at 12:44
• @Edi with rational coefficients? – tarit goswami Oct 4 '18 at 12:44
• If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots. – GEdgar Oct 4 '18 at 12:45
• There isn't a clear, universally accepted, convention on "irrational $\implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent. – lulu Oct 4 '18 at 12:51

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

• When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only. – GEdgar Oct 4 '18 at 12:58
• @GEdgar: that's quite right. Of course, you can infer which case is targeted here. – Yves Daoust Oct 4 '18 at 12:58
• @GEdgar When I posted my answer, there was only one question. – José Carlos Santos Oct 4 '18 at 13:13
• Adding my 2nd question was almost simultaneous with your answer! :) – user231343 Oct 4 '18 at 14:17

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}

The last one was built from

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

so that the roots are

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

• Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2). – Calvin Godfrey Oct 4 '18 at 13:29

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 $$q$$)

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.

• two rationals one irrational, is my question – user231343 Oct 4 '18 at 12:51
• This case is already known by the OP. – Yves Daoust Oct 4 '18 at 12:52
• Note $0$ is rational. So it is better to use the $x^2$ coefficient. – GEdgar Oct 4 '18 at 12:56
• @GEdgar You are correct, I fixed it. – 5xum Oct 4 '18 at 13:05

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.

• See José's answer for three irrational roots with sum $0$ and product $-1$. – GEdgar Oct 4 '18 at 13:02