<s>[INST] 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?
 [/INST] 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. 
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