Algebra 2
posted by Vic .
Factor f(x)=x^4+x^38x^2+6x+36 completely

If there is a rational root, it would be +or 1,2,3,4,6,8 or 12.
x = 2 works, so x+2 is a factor.
So, after one factoring step, and a bit of polynomial long division, you have
(x+2)(x^3 x^2 6x +18) = 0
Next, use the rational root theorm on the cubic factor. You will find that 3 is a root, making (x+3) a factor.
So now you have
x^4+x^38x^2+6x+36
(x+2)(x+3)(x^2 4x +6)
The remaining two factors are complex conjugates.