Is $x+2$ invertible in given quotient ring? 
Given the quotient ring $R=\mathbb{Q}[x]/(x^2-7)$. Is $x+2$ invertible in $R$? 

I guess it is, but how to find it?
 A: Hint:
$$\frac{1}{\sqrt{7}+2}=\frac{1}{\sqrt{7}+2}\cdot\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{-2}{3}+\frac{\sqrt{7}}{3}$$
A: Hint: Use division with remainder to find $q, r$ such that $x^2 - 7 = q(x)(x + 2) + r(x)$. What can you say about the degree of $r$?
A: Hint:
$x^2-7$ and $x+1$ don't have any common roots, so they are coprime in $\Bbb Q[x]$. This means you can use the extended Euclidean algorithm to find $a(x)$ and $b(x)$ such that $1=a(x)(x^2-7)+b(x)(x+1)$. What does this equation look like in your ring?
A: $\, y:=x\!+\!2,\,\ 0 = x^2\!-7 = (y\!-\!2)^2\!-7 = y(y\!-\!4)-3\,\Rightarrow\,  1/y = (y\!-\!4)/3 = (x\!-\!2)/3\ \ $ QED
Similarly, over a field,  one can use a shift to transform a nonzero polynomial with $\,r\,$ as a root, into a polynomial having $\,s = r+a\,$ as a root. Any such polynomial yields the inverse, since
$$\ 0 = f(s) = s g(s) - f_0\ \Rightarrow\ s g(s) = f_0\ \Rightarrow \dfrac{1}s = \dfrac{g(s)}{f_0}$$
Note that we can always assume $\, f_0 = f(0)\ne 0\,$ by cancelling all factors of $\,x\,$ from $\,f(x).\ $ This can be considered a generalization of rationalizing denominators (as in Zev's post), which uses the fact that an algebraic number divides its norm (a special case of the above, when $\, f_0 =$ norm).
A: Thanks you all. In the end I used this idea:
If it is invertible then there are $a,b \in \mathbb{Q}$ such that:
$$(ax+b)(x+2)=1$$
$$ax^2+2ax+bx+2b=1$$
Then use that $x^2=7$ 
$$(2a+b)x+7a+2b=1$$
From her we get a system of equations:
$$\left\{ 
  \begin{array}{l }
    2a+b=0\\
    7a+2b=1
  \end{array} \right.$$
And after finding $a$ and $b$ we can express:
$$(x+2)^{-1}=ax+b=\dfrac{x}{3}-\dfrac{2}{3}$$
