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This question already has an answer here:

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?

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marked as duplicate by lulu, Winther, Holo, Namaste algebra-precalculus Oct 4 '18 at 16:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @ChinnapparajR, two rationals one irrational, is my question $\endgroup$ – user231343 Oct 4 '18 at 12:44
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    $\begingroup$ Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job. $\endgroup$ – lulu Oct 4 '18 at 12:44
  • $\begingroup$ @Edi with rational coefficients? $\endgroup$ – tarit goswami Oct 4 '18 at 12:44
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    $\begingroup$ If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots. $\endgroup$ – GEdgar Oct 4 '18 at 12:45
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    $\begingroup$ 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. $\endgroup$ – lulu Oct 4 '18 at 12:51
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Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.

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    $\begingroup$ 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. $\endgroup$ – GEdgar Oct 4 '18 at 12:58
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    $\begingroup$ @GEdgar: that's quite right. Of course, you can infer which case is targeted here. $\endgroup$ – Yves Daoust Oct 4 '18 at 12:58
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    $\begingroup$ @GEdgar When I posted my answer, there was only one question. $\endgroup$ – José Carlos Santos Oct 4 '18 at 13:13
  • $\begingroup$ Adding my 2nd question was almost simultaneous with your answer! :) $\endgroup$ – user231343 Oct 4 '18 at 14:17
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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

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

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

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

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