x^4+3x^2+4 has no linear factors with Real coefficients since x^4+3x^2+4 >= 4 for all x in RR.
So let's look for quadratic factors. Since there's no term in x^3, these must take the form:
x^4+3x^2+4 = (x^2+ax+b)(x^2-ax+c)
=x^4+(b+c-a^2)x^2+a(c-b)x+bc
Equating coefficients we find:
(i) b+c-a^2 = 3
(ii) a(c-b) = 0
(iii) bc = 4
From (ii), a = 0 and/or b = c.
If a = 0, then we are effectively trying to find linear factors of t^2+3t+4, where t = x^2. This quadratic in t has a negative discriminant (3^2-4*1*4 = 9-16 = -7), so only Complex zeros.
So try b=c:
From (iii), we get b = c = 2 or b = c = -2. If a in RR then a^2 >= 0 and from (i) we get b+c = 3+a^2 >= 3. So b = c = 2.
Then a^2 = b+c-3 = 2+2-3 = 1. So a = +-1
Hence x^4+3x^2+4 = (x^2+x+2)(x^2-x+2)
