Solve $3^{4x}-3^{3x}-7\times 3^{2x}+ 3^x + 6 < 0$ I'd like to solve the following inequality $3^{4x}-3^{3x}-7\times 3^{2x}+ 3^x + 6 < 0$.
I made it so that
$$y = 3^{x}$$
I then replaced $3^{x}$ with $y$:
$$y^4-y^3-7y^2+y+6<0$$
I then used Ruffini's rule to break it down a little bit more:
$$y^4-y^3-7y^2+y+6<0 \rightarrow (y+1)(y^3-2y^2-5y+6)<0$$
Though, I'm not sure using Ruffini's rule was the best approach to this problem and the result I got from isn't probably correct because $(y^3-2y^2-5y+6)$ seems to be having complex numbers as solutions. Any hints on what I could have done better?

Notice that by rational root theorem , the roots of 
$p(z)=z^4-z^3-7z^2+z+6$ are $z=-2,-1,1,3$
Hence the polynomial can be factored into
$p(z)=(z-1)(z-3)(z+2)(z+1)$
Where $z=3^x$
when $p(z)<0$
We have
$$(3^x-1)(3^x-3)(3^x+2)(3^x+1)<0$$
However $3^x$ is strictly increasing and is never negative 
So we have $3^x=1 \Leftrightarrow x=0 $ and $3^x=3^1\Leftrightarrow x=1$
Hence the inequality is just $0<x<1$
