An element does not belong to an ideal 
How can I prove that the element $x-5$ does not belong to the ideal $(x^2-25,-4x+20)$ in $\mathbb Z[x]$. 

I tried to show that by proving $x-5\neq(x^2-25)f(x)+(-4x+20)g(x)$ for all $f,g$. Any help?
 A: Take the formal derivative of $$x-5=(x^2-25)f(x)+(-4x+20)g(x) $$
to arrive at 
$$1 = 2xf(x)+(x^2-25)f'(x)-4g(x)+(-4x+20)g'(x).$$
The right hand side is even at $x=1$.
A: Evaluate $x-5=(x^2-25)f(x)+(-4x+20)g(x)$ at $x=-5$ to obtain $40\mid10$, impossible.
A: $\begin{eqnarray}{\bf Hint}\ \ \ x-5\, &=&\, (x^2\!-25) f(x) -4(x-5) g(x)\\ 
\overset{\phantom{I}}\iff 1\, &=&\, (x\ +\ 5)\ f(x) - 4\, g(x)\\
\overset{x\ =\ -5}\Rightarrow\  1\, &=&\, -4\ g(-5)\ \ {\rm in}\ \ \Bbb Z\ \Rightarrow\Leftarrow
\end{eqnarray}$
A: Compute in the quotient ring $Q:=\mathbb{Z}[x]/(x^2-25,-4x+20) \cong (\mathbb{Z}[x]/(x^2-25))/(4x-20)$. Since $x^2-25$ is monic of degree $2$, $\mathbb{Z}[x]/(x^2-25)$ is a free $\mathbb{Z}$-module with basis $1,x$. We divide out the ideal generated by $4x-20$, i.e. the subgroup generated by $4x-20$ and $x(4x-20)=4x^2-20x=4 \cdot 25 - 5 \cdot 4x=5(20-40x)$, i.e. this generator becomes superfluous. Clearly $x-5 \notin \mathbb{Z} \cdot (4x-20)$, so that $x-5 \neq 0$ in $Q$. In fact we can determine the structure of $Q$ as a $\mathbb{Z}$-module: It is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}/4$ with generators $1,x-5$.
This also offers another simple proof: $Q/(2) = \mathbb{F}_2[x]/(x^2-1)$ and here $x-5=x-1 \neq 0$ is clear. (see my comment to Hagen's answer for a quotient-free version of this argument).
A: ${\rm mod}\ \color{#c00}2\!:\  (x^2\!-25) f -\color{#c00}2\, g\ $ is either $\,0\,$ or of degree $\ge 2,\,$ so is $\,\not\equiv\, x-5$
