Answer
$3y(x^2+4y^2)(x+2y)(x-2y)$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $3x^4y-48y^5=\\=3y(x^4-16y^4)\\=3y((x^2)^-(4y^2)^2)\\=3y(x^2+4y^2)(x^2-4y^2)\\=3y(x^2+4y^2)(x^2-(2y)^2)\\=3y(x^2+4y^2)(x+2y)(x-2y)$
