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Find a polynomial of degree 3 that has three real zeros, only one of which is rational.

My answer: $(x - \sqrt{2})(x - 3)(x - \pi)$.

Is this correct? It does have two irrational zeros, but I'm not sure if I'm 100% right.

P.S. Can I use a similar technique to come with an expression for the following question: A polynomial of degree 4 that has four real zeros, none of which is rational?

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    $\begingroup$ That'll do it. To make it harder you might try to find an example that has integer coefficients. $\endgroup$ – lulu May 28 '16 at 14:15
  • $\begingroup$ Can I just expand this expression and then multiply throughout to get the result? $\endgroup$ – MathEnthusiast May 28 '16 at 14:17
  • $\begingroup$ No, you can't. $\pi$ is transcendental, so will never be the root of a polynomial with integer coefficients. $\endgroup$ – lulu May 28 '16 at 14:18
  • $\begingroup$ @user331377 Yes that would give a polynomial satisfying the question (but not lulu's supplementary question), but it would be a lot less effort to pick something like $(x-1)(x^2-2)=x^3-x^2-2x+2$ $\endgroup$ – almagest May 28 '16 at 14:19
  • $\begingroup$ How about replacing $\pi$ with another irrational number like $\sqrt{2}$? $\endgroup$ – MathEnthusiast May 28 '16 at 14:19
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Yes, your answer is correct.

Another way to come up with some is to use the form $x^3-nx \;\;\forall\; n\in\Bbb N \land \sqrt n \notin \Bbb N$ (i.e. where $n$ isn't a perfect square but natural).

For the quartic, consider the equation $x^4-(a+b)x^2+(ab)\;\;\forall \;a,b \in \Bbb N \land \sqrt a, \sqrt b \notin \Bbb N$ (i.e. where $a+b$ is the sum of two non-perfect square natural numbers and $ab$ is their product).

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  • $\begingroup$ I got the part about my answer being correct, but the second one... If I take something like $x^3 - 5x$, how do I know that it has only one rational zero? $\endgroup$ – MathEnthusiast May 28 '16 at 14:25
  • $\begingroup$ $x^3-5x=x(x^2-5)=x(x+\sqrt5)(x-\sqrt5)$ Ergo, $x=0,\pm\sqrt5$ $\endgroup$ – Lanier Freeman May 28 '16 at 14:26
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    $\begingroup$ Ah, got it! I forgot to factor first! $\endgroup$ – MathEnthusiast May 28 '16 at 14:29
  • $\begingroup$ You're very welcome. $\endgroup$ – Lanier Freeman May 28 '16 at 14:36
  • $\begingroup$ Did you mean quartic? $\endgroup$ – user1008646 May 29 '16 at 1:23
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For your question at the end, yes, a similiar technique would work.

If you know $n$ zeroes of a polynomial of degree $n$, then you know all of them, because there are at most $n$ of them. So if you know four irrational zeroes, you know that there are no rational zeroes. In particular, for a degree 4 polynomial, you know that there are no rational zeroes if you know four irrational zeroes.

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Very simple example with integer coefficients:

$(x^2-2)(x-1) =x^3-x^2-2x+2 $.

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