Show $I=\{sf\mid s\in \mathbb Z_{11}[x]\}$ matches $J=\{h \in \mathbb Z_{11}[x] \mid h(1) = h(-2) = 0\}$. Let $f=x^2+x+\overline{9} \in \mathbb Z_{11}[x]$. Show that $I=\left\{sf\mid s\in \mathbb Z_{11}[x] \right\}$ matches $J=\left\{h \in \mathbb Z_{11}[x] \mid h(\overline{1}) = h(\overline{-2}) = \overline{0}\right\}$. I don't have the palest idea where to start, can you please push me in the right direction?
 A: Note that $F=\mathbb{Z}_{11}$ is a field (although actually, the only thing you need is that $-3=8$ is invertible.
Then recall that for a polynomial $f\in F[X]$ and $a\in F$
$$
f(a)=0\quad\Leftrightarrow \quad f(x)=(x-a)g(x)
$$
for some $g\in F[X]$.
Apply this to $f\in J$ and $a=1$. This gives 
$$
f(x)=(x-1)g(x).
$$
Now $f(-2)=0=-3g(-2)$ so $g(-2)=0$ since $-2-1=-3\neq 0$ is invertible in $F$. So
$$
g(x)=(x+2)h(x).
$$
Finally
$$
f(x)=(x-1)(x+2)h(x)=(x^2+x-2)h(x)=(x^2+x+9)h(x).
$$
So $J$ is contained in $I$. And the converse is trivial.
A: Can you see how every element $x\in I$ has $x(\overline 1)=x(-\overline 2)=0$? If $\overline 1$ and $-\overline 2=\overline 9$ are zeros of $h\in J$, can you give a factor of $h$?
A: Hint $\ $  Notice $\rm\ mod\ 11\!:\ x^2\!+x+9 \,\equiv\, (x-1)(x+2),\ $ then apply the  following
Bifactor Theorem $\ $ Let $\rm\,a,b\in R,\,$ a ring, and $\rm\:f\in R[x].\:$ If $\rm\ \color{#C00}{a\!-\!b}\ $ is cancelable in $\rm\,R\,$ then 
$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in R[x]$$
Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying  Factor Theorem twice, canceling $\rm\: \color{#C00}{a\!-\!b},\:$ we obtain 
$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in R[x]\\
\rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in R[x]\\
&\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$
Remark $\ $ The theorem  may fail when $\rm\ a\!-\!b\ $ is not cancelable (i.e. is a zero-divisor), e.g.
$\rm\quad mod\ 8\!:\,\ f(x)=x^2\!-1\,\Rightarrow\,f(3)\equiv 0\equiv f(1)\ \ but\ \  x^2\!-1\not\equiv (x\!-\!3)(x\!-\!1)\equiv x^2\!-4x+3$ 
