Question

x^4+3x^2+4 factorize ​

Answer 1

Explanation:

x4+3x2+4 has no linear factors with Real coefficients since x4+3x2+4≥4for all x∈R.
So let's look for quadratic factors. Since there's no term in x3, these must take the form:
x4+3x2+4=(x2+ax+b)(x2−ax+c)
=x4+(b+c−a2)x2+a(c−b)x+bc
Equating coefficients we find:
(i) b+c−a2=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 t2+3t+4, where t=x2. This quadratic in t has a negative discriminant (32−4⋅1⋅4=9−16=−7), so only Complex zeros.
So try b=c:
From (iii), we get b=