Let $P_3(\mathbb{R})$ be the vector space of real polynomials of degree ≤2.
Now let $\lambda\in \mathbb{C}\backslash\mathbb{R}$ be a complex number, which is not a real number.
1) Show that for all $p,q \in P_3(\mathbb{R})$ that:
$$p(\lambda)\overline{q(\lambda)}+p(1)q(1)+\overline{p(\lambda)}q(\lambda) $$
is a real number.
2) Show that a polynomial $p\in P_3(\mathbb{R})$, which is not $0$, cannot have both $1$ and $\lambda$ as roots.
1) I can show that it is a real number by taking the conjugate of the entire equation and prove that it does not change: $\overline{p(\lambda)\overline{q(\lambda)}+p(1)q(1)+\overline{p(\lambda)}q(\lambda)} = \overline{p(\lambda)}q(\lambda)+p(1)q(1)+p(\lambda)\overline{q(\lambda)}=p(\lambda)\overline{q(\lambda)}+p(1)q(1)+\overline{p(\lambda)}q(\lambda) $
Which means that it must be a real number as it does not change under complex conjugation, is it correct?
2) I hope someone can show how to this part or at least give a hint as to how it should be done.
If $1$ and $\lambda$ are roots, then $\overline\lambda$ is also a root and, since $\lambda\notin\mathbb R$, you would have a non-null polynomial with degree less than $3$ with $3$ distinct roots, which is impossible.