Let $(\Bbb R[x],+,.)$ a ring and $S=${$p \in \Bbb R[x] : p(1) = 0 $}. $(S,+,.)$ $⊂$ $\Bbb R[x]$. Let's see that $S$ is a subring of $(\Bbb R[x],+,.)$
(1) $S$ is nonempty: Let $p(x)=a(x)(x-1)$ with $a(x) \in \Bbb R[x]$ then $p(1)=a(1)(1-1)=0 \in S$ so $S$ is nonempty.
(2) Let $p,q \in S$ then $p(1)=0$ and $q(1)=0$. Then, I can write $p$ and $q$ as follows: $p(x) = a(x)(x-1)$ and $q(x)=b(x)(x-1)$ with $a(x),b(x) \in \Bbb R[x]$.
Then $p(x)+q(x)=t(x)= [a(x)(x-1)]+[b(x)(x-1)]=[a(x)+b(x)](x-1)$ $\Rightarrow$ $t(1)=[a(1)+b(1)](1-1)=0.$
So $p+q \in S$
Now, $p(x).q(x)=r(x)=[a(x)(x-1)].[b(x)(x-1)]=[a(x).b(x)](x-1)^2$ $\Rightarrow$ $r(1)=[a(1).b(1)](1-1)^2=0$
So $p.q \in S$
(3) Let $p \in S$ then I can write p as follows: $p(x)=a(x)(x-1)$ with $a(x) \in \Bbb R[x]$. Let $l$ the opposite of $p$ $\Rightarrow$ $l(x)=-a(x)(x-1)$ $\Rightarrow$ $l(1)=-a(1)(1-1)=0$
So $(S,+,.)$ is subring of $(\Bbb R[x],+,.)$
Is it correct and is it well demonstrated what I did?