$\displaystyle 4(2^x)+31(2^x)-8=0$

OK, I have a good feeling it's going to turn into a complex trinomial.

By substituting $\displaystyle (2^x)$ with a variable, say "z"

That would give us..

$\displaystyle 4z+31z-8=0$

Still not a trinomial..hmm

I see that the factor may be, -1 and 32.

Or, do I somehow break down the 4 into $\displaystyle 2^2$. But still the exponent 2 would not be over the z.