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)#
