#15x^4+11x^2-12 = 15(x^2)^2+11(x^2)-12#
is of the form #ay^2+by+c#, with #y=x^2#, #a=15#, #b=11# and #c=-12#.
The discriminant of this quadratic is given by the formula:
#Delta = b^2-4ac#
#= 11^2-(4xx15xx(-12)) = 121 + 720 = 841 = 29^2#
...a perfect square positive number, so #15y^2+11y-12 = 0# has two distinct rational roots, given by the formula
#y = (-b +-sqrt(Delta))/(2a) = (-11+-29)/30#
That is #y = 18/30 = 3/5# and #y = -40/30 = -4/3#.
These roots give us two corresponding factors #(5y-3)# and #(3y+4)#.
So #15y^2+11y-12 = (5y - 3)(3y + 4)#
Substituting #y=x^2# into this, we get
#15x^4+11x^2-12 = (5x^2 - 3)(3x^2 + 4)#
If we allow irrational coefficients then the first quadratic factor can by broken down further to give us:
#15x^4+11x^2-12 = (sqrt(5)x + sqrt(3))(sqrt(5)x - sqrt(3))(3x^2 + 4)#
This is an instance of #(A^2-B^2) = (A+B)(A-B)#
That's as far as we can go with real coefficients since #3x^2+4 > 0# for all real values of #x#, but with complex coefficients we can factor the second quadratic factor, giving:
#15x^4+11x^2-12#
#= (sqrt(5)x + sqrt(3))(sqrt(5)x - sqrt(3))(sqrt(3)x + 2i)(sqrt(3)x-2i)#
