Finding zero divisors in a polynomial quotient ring 
Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?

I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor. 
What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangle)(g(x)+\langle x^3+5x^2+2x+5\rangle)=\langle x^3+5x^2+2x+5\rangle$$ in the quotient ring.
I know how to find an inverse, is this any similar? How is it done?

$ f=x(x\!+\!1)$ is a $0$-divisor mod $g\!\iff\!  d:=\gcd(f,g)\neq 1,g$ $\!\iff\!  g(0)=0\,$ xor $g(-1)=0\,$ 
Then we have $\!\bmod g\!:\ (\color{#c00}{g/d})f \equiv (f/d)g\equiv  0,\,$ both $\,g/d,f\not\equiv 0.\,$ Thus you seek $\,\color{#c00}{g/d}$
This is true in the OP: $\,g(0)= 5\neq 0,\,\  g(-1)=7=0\,$ in $\Bbb Z_{\large 7},\ $ hence $\,d = x\!+\!1$
