A polynomial of degree 3 that has three real zeros, only one of which is rational.
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?
$\endgroup$ 153 Answers
$\begingroup$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).
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ $\begingroup$Very simple example with integer coefficients:
$(x^2-2)(x-1) =x^3-x^2-2x+2 $.
$\endgroup$
