Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded: 
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x + 7) + 8] + 15\\
& = (x^2 + 8x + 7)^2 + 8(x^2 + 8x + 7) + 15
\end{align*}
If we make the substitution $y = x^2 + 8x + 7$, we get
\begin{align*}
y^2 + 8y + 15 = (y^2 + 3y) + (5y + 15) = y(y+3) + 5(y+3) = (y+5)(y+3) = 0
\end{align*}
From whence we obtain that:
\begin{align*}
y + 5 = 0\Leftrightarrow x^2 + 8x + 12 = 0 \Leftrightarrow (x+4)^2 - 4 = 0\Leftrightarrow x\in\{-6,-2\}\\
\end{align*}
Analogously, we have that
\begin{align*}
y + 3 = 0\Leftrightarrow x^2 + 8x + 10 = 0\Leftrightarrow (x+4)^2 - 6 = 0\Leftrightarrow x\in\{-4-\sqrt{6},-4+\sqrt{6}\}
\end{align*}
Finally, the solution set is given by $S = \{-6,-2,-4-\sqrt{6},-4+\sqrt{6}\}$.
Differently from this approach, could someone provide me an alternative way of solving this problem? Any contribution is appreciated. Thanks in advance.

HINT.-Looking about integer solutions for $f(x)=(x+1)(x+3)(x+5)(x+7) + 15 = 0$  possible values should be even and negative so the only candidates are $-2,-4$ and $-6$. We verified that $-2$ and $-6$ are roots. The other two roots are solutions of $$\frac{x^4+16x^3+86x^2+176x+120}{x^2+8x+12}=x^2+8x+10$$
