Remainder Theorem Problem 
When the polynomial $p(x)$ is divided by $(x^2+3x+2)$ the remainder is $5x+1$.  Find the remainder when $p(x)$ is divided by $(x+2)$.

I get that $(x+2)$ is a factor of $(x^2+3x+2)$ but I'm not sure how to use that fact to get the question's answer.
 A: There is polynomial $q$, for which:
$$p(x)=q(x)(x+1)(x+2)+5x+1=q(x)(x+1)(x+2)+5x+10-9,$$
which gives the answer: $$-9$$
A: Write $$p(x)= q(x)(x+1)(x+2)+5x+1$$ Put $x=-2$ and you get $p(-2) = -9$. So remainder when $p(x)$ is divided by $x+2$ is $-9$. 
A: Think of it this way:
$p(x)$ divided $x^2 + 3x + 2$ has remainder $5x+1$ means:
$$p(x) = q(x)(x^2 + 3x + 2) + 5x+1$$ where $q(x)$ is a polynomial.
But we have $x^2 + 3x + 2$ so
$$p(x) = q(x)(x+1)(x+2) + 5x + 1$$
$$= s(x)(x+2) + 5x+1$$
Where $s(x) = q(x)(x+1)$
So we just need the divide $5x + 1$ by $x+2$ to get $5x + 1 = t(x)(x+2) + r(x)$.
As $5x + 1 = 5(x+2) - 9$ we have
$$p(x) = q(x)(x+1)(x+2) + 5(x+2) - 9$$
$$= [q(x)(x+1) + 5](x+2) - 9$$.
And the remainder is $9$.
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This is actually the exact same logic as it would be with numbers.
If $M$ divided by $36$ has remainder $17$, then what is the remainder of $M$ divided by $12$.
As $12$ divides $36$ we know that $12$ will divide "just as evenly" as $36$ and the remainder will be the same $17$ but further divided by $12$ to have remainder: $5$.
Formally:  $$M = k*36 + 17$$
$$= (3k)*12 + (12 + 5)$$
$$= (3k+1)*12 + 5$$.
The remainder is $5$ because the remainder of $17$ is $5$.
It's no different with polynomials.
There remainder is $-9$ because the remainder of $5x + 1$ is $-9$.
A: $$\ \ \begin{align}p\ &\equiv\,\ 5\,\ x\,\  +\  1\!\!\pmod{\!(x + 2)f}\quad\ \ [\,f = x + 1\rm \ \ in\ OP\,]\\
\Rightarrow\ \ p\ &\equiv\,\ 5\ \,\color{#c00}x\,\ +\ 1\!\!\pmod{\color{#0a0}{x+2}}\\
&\equiv\, 5(\color{#C00}{-2})\!+ 1\quad  {\rm by}\quad\, \ \color{#0a0}{x + 2}\equiv 0\,\Rightarrow\, \color{#c00}{x\equiv -2}
\end{align}$$
Remark $ $ Sometimes authors write such congruences as  $\!\!\pmod{\color{#c00}{x\equiv -2}}\, $ to make such evaluations clearer.  The 2nd line uses the fact that congruences always persist mod factors of the modulus by $$\,p\equiv q\!\!\pmod{\!mn}\,\Rightarrow\, \underbrace{m\mid mn\mid p-q\,\Rightarrow\, m\mid p-q}_{\large\text{$\it{transitivity}$ of divisibility}}\,\Rightarrow\,p\equiv q\!\!\pmod{\!m}$$ 
