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)
