Show that a polynomial p can't have both roots 1 and i I have to show that a polynomial $p\in P_3(\mathbb{R})$ where $p\neq0$, can't have both roots, $1$ and $i$, where $i^2=-1$. The polynomial has degree 2 or less.

$p(\alpha)=c_0+c_1\alpha+c_2\alpha^2$

I know that, when we want to find the roots of a function, we have to solve the function for $x$ so that $f(x)=0$

I don't know how to solve this kind of question, when $p$ cant be equal to $0$ and I have to show that the polynomial $p\in P_3(\mathbb{R})$ can't have both roots 1 and $i$.

Thanks in advance
If we have a polynomial with real coefficients, with $z$ as a root, one can check that $\overline{z}$ must also be a root of the polynomial. Thus, if $i$ is a root of a polynomial $p$, then so is $-i$. If we also require that $1$ be a root of $p$, then $p$ has at least $3$ roots. But how many roots can $p$ have, given that it's of degree at most $2$?
I don't understand the fuss... Just write $p(x)=(x-1)(x-i)$ and expand it to get $p(x)=x^2-(1+i)x+i$, which clearly does not belong to $P_{3}(\mathbb{R})$. Or is it?