Factoring a 4th degree trinomial I am trying to factor $3x^4-8x^3+16$, but I have no idea how to even start. I put into Wolfram Alpha, and it said that the answer was $(x-2)^2 (3 x^2+4 x+4)$. How would you factor something like this by hand? Is there any way without using the quartic formula in this case?

You can factor it if you know its roots. In general if $r$ is a root of a polynomial $f(x)$, then $(x-r)$ divides it.
For example, if you see that  $f(x)=3 x^4−8 x^3+16$ has the root $2$, that is, $3(2)^4-8(2)^3+16=48-64+16=0$, then you know that $(x-2)$ divides it. You can then use long division to show that $3 x^4−8 x^3+16= (x-2)(3x^3-2x^2-4x-8) $. Now noticing that $2$ is also a root of $3x^3-2x^2-4x-8$ we have that $(x-2)$ divides that also. By long division $3x^3-2x^2-4x-8 = (x-2)(3x^2+4x+4)$. Now $f(x)=(x-2)^2(3x^2+4x+4)$.
So what you can do is first look for some roots using various methods. 
