First notice that 36x^4+15x^2+4 has no linear factors with real coefficients, because 36x^4 >= 0 and 15x^2 >= 0 so
36x^4+15x^2+4 >= 4 > 0 for all x in RR.
Suppose it has quadratic factors:
36x^4+15x^2+4 = (6x^2+ax+b)(6x^2+cx+d)
= 36x^4+6(a+c)x^3+(6d+6b+ac)x^2+(ad+bc)x+bd
Comparing the coefficients of x^3 we find c=-a so this simplifies to:
= 36x^4+(6d+6b-a^2)x^2+a(d-b)x+bd
Looking at the coefficient of x, either a=0 or d = b.
If a=0 then 6(b+d) = 15 and bd=4. Substituting d=4/b in 6(b+d)=15 we get:
6(b+4/b)=15, and hence 6b^2-15b+24=0. This has no real solutions, since it's discriminant is 225-576=-351.
How about the other possibility: d = b?
Then b=d=+-2 in order that bd=4.
Looking at the coefficient of x^2 we have
15 = (6d+6b-a^2) = 12b - a^2, hence b=2 and a=+-3.
So 36x^4+15x^2+4 = (6x^2+3x+2)(6x^2-3x+2)
