Could someone please verify whether I am missing something in my solution?
Let $F$ be a field and $a\in F$. Show that if $p(x)\in F[x]$ is divided by $x-a$, the remainder is $p(a)$.
By the division algorithm, if $p(x)$ is divided by $x-a$, there exist unique polynomials $q(x),r(x)\in F[x]$ such that $p(x)=(x-a)q(x)+r(x)$ with $degr(x)<deg(x-a)=1$ or $r(x)=0$. Then $r(x)$ is a constant. If $x=a$, then $p(a)=(a-a)q(a)+r(a)$ or $p(a)=r(a)$. Then the remainder is $p(a)$.
I feel I am missing something in my solution and not connecting ideas well.