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?
Best Answer
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).