Show that the set of polynomials with 1 as a root form a linear subspace

Question

Let $\mathbb{C}(x)$ be the vector space $\mathbb{C}$ of polynomials $p\left(x\right)$ in one variable $x$ with coefficients in $\mathbb{C}$.

Is the set $p(x) \in \mathbb{C}\left(x\right)$ such that $1$ is a root of $p(x)$ a linear subspace?

Also show whether the set $p(x) \in \mathbb{C}(x)$ such that $1$ is not a root of $p(x)$ is a linear subspace or not?

So to my understanding, the set such that $1$ is a root of $p(x)$ is all polynomials with $p(-1)=0$ and I understand we need to determine whether these polynomials are closed under addition and scalar multiplication, but I don't know how to do this without the question defining the polynomials degree.

Also for the set such that $1$ is not a root, I assume this is all polynomials with $p(-1)\neq0$

Answer 1

$1$ is a root of $p(x)$ iff $p(1) = 0$. If $p,q$ are two such polynomials $(p+q)(1) = p(1) + q(1) = 0$ and for $\lambda \in \mathbb{C}, \lambda p(1) = \lambda \cdot 0 = 0$. (Note you need to bound $\deg(p+q)$ above also but that is straightforward) Then since your set is non-empty it is a subspace.

The set of all polynomials over $\mathbb{C}$ where $1$ is not a root isn't a subspace. (edit: to clarify this is the set of polynomials $f$ such that $f(1) \neq 0$) Try considering $p(x) - 1$ and $q(x) + 1$ with $p,q$ as before.

Answer 2

Hint Show that the evaluation map $\Bbb C [x] \to \Bbb C$ defined by $1 \mapsto p(1)$ is a linear map. By definition its kernel is the set of polynomials for which $1$ is a root, and the kernel of any linear map is a (linear) subspace.