A good way to describe a kernel of a group homomorphism 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 describe the kernel of $\psi$ as simply as possible. Is there a better way to answer this than just saying that it's the set of all polynomials such that $p(3) = 0$? or $\{ p(x) : p(3) = 0 \}$?
 A: Not really. You can say in words that the kernel is the set of all polynomials that have $3$ as a root, or equivalently all polynomials that contain the linear factor $x-3$ in their factorisation.
In abstract algebra, this is also called the kernel of the evaluation map at the point $3$. It does not get a special name.
A: We first observe that
$\psi: \Bbb R[x] \to \Bbb R, \; \psi(p(x)) = p(3) \tag 0$
is in fact a ring homomorphism, the well-known evaluation homomorphism of $\Bbb R[x]$ at $x = 3$.  Thus, we use some ring-theoretic concepts in the following argument.
We have
$\ker \psi = (x - 3) = \{(x - 3)q(x), \; q(x) \in \Bbb R[x] \}, \tag 1$
where $(x - 3)$ is the principal ideal generated by $x - 3$, that is, the set of multiples of $x - 3$ in $\Bbb R[x]$.  That this is in fact the case is easily seen using Euclidean division of polynomials, which states that any $p(x) \in \Bbb R[x]$ may be written in the form
$p(x) = (x - 3)q(x) + r \tag 2$
with
$\deg r = 0, \tag 3$
i.e.,
$r \in \Bbb R \tag 4$
is constant.  If we evaluate (2) at
$x = 3 \tag 5$
we find
$0 = p(3) = (3 - 3)q(3) + r = 0 + r = r, \tag 6$
whence (2) becomes
$p(x) = (x - 3)q(x), \tag 7$
and clearly, any $p(x) \in \Bbb R[x]$ of this form satisfies
$p(3) = (3 - 3)q(3) = 0q(3) = 0, \tag 8$
thus proving our assertion (1).  Indeed, (1) describes $\ker \psi$ in (relatively) simple mathematical terms.
We note in closing that we may replace $3$ with any $a \in \Bbb R$ in the above and obtain the analogous result
$\ker \psi = (x - a) = \{(x - a)q(x), \; q(x) \in \Bbb [x] \}. \tag 9$
