Factor completely. 9x^4 -225y^8

Question

I'm stumped on how to factor polynomials. I've done a few, but I can't solve this one.

Answer 1

What we have here is the difference of two squares. The general expression is (a^2-b^2) which factorises into (a-b)(a+b). In the given equation a=3x^2 and b=15y^4, so we have initially (3x^2-15y^4)(3x^2+15y^4). But 3 is a common factor to both these factors so can be taken outside the brackets as 3*3=9: (x^2-5y^4)(x^2+5y^4). The factor 9 could have been taken out of the original equation: 9(x^4-25^4). Does this factorise further? If irrational numbers are allowed x^2-5y^4=(x-y^2sqrt(5))(x+y^2sqrt(5)), but the requirement to factorise usually sticks to rational numbers.