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?

The long division of $x^3+5x^2+2x+5$ by the GCD $x+1$ yields $x^2+4x+5$, so we have (in the given ring)
$$(x^2+x)(x^2+4x+5)=x(x+1)(x^2+4x+5)=x(x^3+5x^2+2x+5)=x\cdot0=0$$
Thus if $f(x)=x^2+x$, $g(x)$ may be $x^2+4x+5$.
