Polynomial function of degree $n$ with exactly $k$ roots

Question

If $n-k$ is even, and $\geq 0$, find a polynomial function of degree $n$ with exactly $k$ roots.
I cannot understand this question. Is this not trivial?

I can take $f(x)=x^3$ as the required polynomial function on $\Bbb{R}$ having only one root $0$ and degree is $3$ so in this case $$n-k=3-1=2=\text{ even}.$$ Is this fine or the question is supposed to tell something else?

Answer 1

That solves the problem in a particular case, namely when $n=3$ and $k=1$. It doesn't solve it in the general case.

In the general case, you can take the polynomial$$x(x-1)(x-2)\ldots\bigl(x-(k-1)\bigr)(x^2+1)^{(n-k)/2}.$$

Answer 2

The general answer will be $$(x-a_1)(x-a_2)\ldots(x-a_{k-1})(x-a_{k})\cdot P(x)^{(n-k)/r}.$$

Where $P(x)$ is an irreducible polynomial of degree $r$ over $\mathbb{Q}.$