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?
 A: Your factorization is correct -- as far as it goes (except you have a horribly misused equals sign in the long displayed formula).
But you can factor it further:
$$ y^3-2y^2-5y+6 = (y+2)(y-1)(y-3)$$
I don't know where you got the idea that the roots would be complex.
So among positive $y$ (since $3^x$ is always positive) your inequality holds for $1<y<3$, which gives you $0<x<1$.
A: Hint: You're doing it correctly.
If the cubic factor has a (nonreal) complex root, then it will have the conjugate as a root as well since the coefficients are real. The cubic will always have at least one real root. So if its other roots are nonreal, then it can be written as the product of a linear factor $(y-a)$ and a quadratic factor which is of constant sign (in this case, positive).
Otherwise it will have three real roots.
A: 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 $0<x<1$
A: $${ y }^{ 3 }-2{ y }^{ 2 }-5y+6={ y }^{ 3 }-{ y }^{ 2 }-{ y }^{ 2 }-6y+y+6={ y }^{ 2 }\left( y-1 \right) -y\left( y-1 \right) -6\left( y-1 \right) =\left( y-1 \right) \left( { y }^{ 2 }-y-6 \right) =\left( y-1 \right) \left( y-3 \right) \left( y+2 \right) $$
