Theorem. If $R$ is a commutative ring with identity, then $R[x]$ is a commutative ring with identity.

Proof (structure proof, coefficient-by-coefficient).
We verify the ring axioms using the definitions above.

• Additive structure. Under addition, coefficientwise addition makes $R[x]$ an abelian group:
  associativity and commutativity follow from the same properties in $R$,
  the zero polynomial is the additive identity,
  and $-p$ is defined by $(-p)(x)=\sum (-a_k)x^k$.
• Multiplication is associative. For $p,q,r\in R[x]$, compare the coefficient of $x^k$ in $(pq)r$ and in $p(qr)$.
  Both coefficients are equal to $\sum_{i+j+\ell=k} a_i b_j c_\ell$, using associativity and distributivity in $R$.
• Distributivity. The coefficient of $x^k$ in $p(q+r)$ equals
  $\sum_{i+j=k} a_i (b_j+c_j)=\sum_{i+j=k} a_i b_j + \sum_{i+j=k} a_i c_j$,
  which is the coefficient of $x^k$ in $pq+pr$.
• Identity and commutativity. The constant polynomial $1$ acts as a multiplicative identity.
  If $R$ is commutative, then $a_i b_j=b_j a_i$ implies $pq=qp$ coefficientwise.

Therefore $R[x]$ is a commutative ring with identity. $\square$

Reference: Standard construction of polynomial rings (Judson; Wikipedia polynomial ring).
(Judson PDF,
Wikipedia: Polynomial ring)

Proposition. For $p,q\in R[x]$,
$$\deg(p+q)\le \max\{\deg(p),\deg(q)\}.$$

Proof.
Let $m=\deg(p)$ and $n=\deg(q)$. Then all coefficients of $p$ above degree $m$ are $0$, and similarly for $q$.
So for $k>\max\{m,n\}$, the coefficient of $x^k$ in $p+q$ is $0+0=0$.
Therefore $p+q$ has no nonzero terms above degree $\max\{m,n\}$, proving the inequality. $\square$

Proposition (Degree of a product).
If $R$ is an integral domain and $p,q\in R[x]$ are nonzero, then
$$\deg(pq)=\deg(p)+\deg(q).$$

Proof.
Write $p(x)=a_m x^m+\text{(lower terms)}$ and $q(x)=b_n x^n+\text{(lower terms)}$
with $a_m\ne 0$ and $b_n\ne 0$.
In the product, the coefficient of $x^{m+n}$ is $a_m b_n$.
Because $R$ is an integral domain, $a_m b_n\ne 0$, so $x^{m+n}$ appears with nonzero coefficient.
No term of degree greater than $m+n$ can appear (degrees add when multiplying monomials).
Hence $\deg(pq)=m+n$. $\square$

Reference: Wikipedia — Polynomial ring (degree properties); standard texts.
(Wikipedia: Polynomial ring)

Theorem (Evaluation is a ring homomorphism).
For each $a\in A$, the map $\operatorname{ev}_a:R[x]\to A$ satisfies:
$$\operatorname{ev}_a(p+q)=\operatorname{ev}_a(p)+\operatorname{ev}_a(q),\quad
\operatorname{ev}_a(pq)=\operatorname{ev}_a(p)\,\operatorname{ev}_a(q),\quad
\operatorname{ev}_a(1)=1.$$

Proof.
Additivity follows by substituting $a$ and using coefficientwise addition.
For multiplicativity, expand:
$$\operatorname{ev}_a(pq)=\sum_{k\ge 0}\left(\sum_{i+j=k} a_i b_j\right)a^k
=\sum_{i,j} a_i b_j a^{i+j}
=\left(\sum_i a_i a^i\right)\left(\sum_j b_j a^j\right)=\operatorname{ev}_a(p)\operatorname{ev}_a(q).$$
Finally $\operatorname{ev}_a(1)=1$. $\square$

Reference: Wikipedia — Polynomial ring (evaluation homomorphism/universal property).
(Wikipedia: Polynomial ring)

Proposition. If $K$ is an infinite field and $p,q\in K[x]$ define the same function $K\to K$,
then $p=q$ as polynomials.

Proof. If $p$ and $q$ define the same function, then $(p-q)(a)=0$ for all $a\in K$.
If $p\ne q$, then $p-q$ is a nonzero polynomial and therefore has only finitely many roots
(at most its degree; see Section “Roots and multiplicity”). This contradicts the fact that $K$ is infinite.
Hence $p-q=0$ and $p=q$. $\square$

Reference: Standard consequence of the “at most $n$ roots” theorem; see Judson polynomials chapter.
(Judson PDF)

Theorem (Degree bounds the number of roots).
If $F$ is a field and $0\ne f\in F[x]$ has degree $n$, then $f$ has at most $n$ roots in $F$.

Proof (induction on $n$).
If $n=0$, then $f$ is a nonzero constant and has no roots.
Assume the statement holds for degrees $<n$.
If $f$ has no roots, we are done. Otherwise let $a$ be a root.
By the Factor Theorem, $f(x)=(x-a)g(x)$ for some $g\in F[x]$ with $\deg(g)=n-1$.
Any root $b\ne a$ of $f$ must satisfy $0=f(b)=(b-a)g(b)$.
Since $b\ne a$, the factor $(b-a)\ne 0$, so $g(b)=0$.
Thus, all roots of $f$ other than $a$ are roots of $g$.
By induction, $g$ has at most $n-1$ roots, so $f$ has at most $1+(n-1)=n$ roots. $\square$

Reference: Standard theorem in the polynomials chapter of abstract algebra texts (Judson).
(Judson PDF)

Proposition (Multiplicity parity and sign change over $\mathbb{R}$).
Let $f\in\mathbb{R}[x]$ and let $a\in\mathbb{R}$. Suppose $a$ is a root of $f$ of multiplicity $m\ge 1$.
Then there exists a polynomial $h\in\mathbb{R}[x]$ with $h(a)\ne 0$ such that
$$f(x)=(x-a)^m h(x).$$
Then $f$ changes sign as $x$ passes through $a$ if and only if $m$ is odd.

Proof.
Because $h$ is a polynomial, it is continuous on $\mathbb{R}$.
Since $h(a)\ne 0$, set $\varepsilon=\frac{1}{2}|h(a)|$.
By continuity, there exists $\delta>0$ such that
$$|x-a|<\delta \ \Longrightarrow\ |h(x)-h(a)|<\varepsilon.$$
For such $x$, we have $|h(x)-h(a)|<|h(a)|$, which forces $h(x)$ to have the same sign as $h(a)$ (in particular, $h(x)\ne 0$).

Now take any $x$ with $0<|x-a|<\delta$.
If $x\in(a-\delta,a)$ then $(x-a)$ is negative, so $(x-a)^m$ is negative exactly when $m$ is odd.
If $x\in(a,a+\delta)$ then $(x-a)$ is positive, so $(x-a)^m$ is positive for every $m$.
Since $h(x)$ keeps a fixed sign near $a$, the product $f(x)=(x-a)^m h(x)$ changes sign from the left side of $a$ to the right side precisely when $m$ is odd. $\square$

Reference: sign-preserving property of continuous functions at a nonzero point (analysis) and the factorization
$f(x)=(x-a)^m h(x)$ for a root of multiplicity $m$ (algebra).
(Hunter, Introductory Real Analysis, Ch. 7;
OpenStax, Sec. 5.3.)