Let $\Bbb R [x]$ denote the group of all polynomials with real coefficients, under the operation of addition. Let $\psi : \Bbb R[x] \to \Bbb R$ be the function defined by $\psi (p(x)) = p(3)$, for all $p(x) \in \Bbb R [x]$.
I'm trying to prove this to be a group homomorphism and so far this is what I got...
Proof:
First, we must show that $\psi (p(x+y)) = \psi (p(x)) + \psi (p(y))$ for all $x,y \in \Bbb R$. Therefore, $(x+y)(3) = 3x + 3y = \psi (p(x)) + \psi (p(y)) = \psi (p(x+y))$, proving that $\psi$ is a group homomorphism.
Just wondering if I have everything I need in this proof for it to be correct.
You’re trying to prove the wrong thing. What you must prove is that if $p,q\in\Bbb R[x]$ (i.e., $p$ and $q$ are polynomials in the indeterminate $x$ with real coefficients), then $$\psi(p+q)=\psi(p)+\psi(q)\;.$$ And this is true:
$$\begin{align*} \psi(p+q)&=(p+q)(3)\\ &=p(3)+q(3)\\ &=\psi(p)+\psi(q)\;. \end{align*}$$