The question is as follows:
Prove that the set of all polynomials in $x$ with nonnegative coefficients and constant term 1 is convex.
Is this just meaning that it follows the form $\lambda x + (1- \lambda)y$, where they are replaced with polynomials? I think I'm just having a difficult time picturing what I need to do
On
The space of all polynomials $P$ is convex. The map $C_k$ that gives the $k$th coefficient of a polynomial is linear.
Hence the subset $\{ p | C_0 p = 1, C_k p \ge 0 ,k > 0\} $ is a convex set.
Alternatively:
Suppose $p(x) = \sum_k p_k x^k$ and $r(x) = \sum_k r_k x^k$ are polynomilals with $r_k,p_k \ge 0 $ and $p_0=r_0 = 1$.
Choose $t \in [0,1]$ and note that the $x^k$ coefficient of $tp+(1-t)r$ is $t p_k + (1-t) r_k$.
It follows that $t p_k + (1-t) r_k \ge 0$ and $t p_0 + (1-t) r_0 = 1$, hence the set is convex.
Let $0 <\lambda <1$. If $p(x)= \sum\limits_{k=0}^{n} a_kx^{k}$ and $q(x)= \sum\limits_{k=0}^{m} b_kx^{k}$ with $n \leq m$ then then $\lambda p(x)+(1-\lambda ) q(x)= \sum\limits_{k=0}^{n} (\lambda a_k +(1-\lambda) b_k)x^{k}+ \sum\limits_{k=n+1}^{m} (1-\lambda) b_k x^{k}$. This is a polynomial of the same type.
The second sum here is absent when $n=m$.