Factor the following expression completely: 12y^5-34xy^4+14x^2y^3

Please show all of your work.

2 * y^3 * (6y^2 - 17xy + 7x^2)

factor 6y^2 - 17xy + 7x^2 with the quadratic formula

y = (17x +/- sqrt(289x^2 - 4 * 6 * 7x^2)) / (2 * 6)
y = (17x +/- sqrt(289x^2 - 168x^2)) / 12
y = (17x +/- sqrt(121x^2)) / 12
y = (17x +/- 11x) / 12
y = 28x/12 , 6x/12
y = 7x / 3 , x / 2

2 * y^3 * (3y - 7x) * (2y - x)

12y^5-34xy^4+14x^2y^3= y^3(12y^2-34xy+14x^2)

12y^2-34xy+14x^2=(7x-3y)(2x-4y) .
=2*(7x-3y)(x-2y)
[hint:consider the equation as a quadratic in x , then solve using formula or factorise]
therefore
12y^5-34xy^4+14x^2y^3
=2*y^3(7x-3y)(2x-4y)

12y^5-34xy^4+14x^2y^3
= 2y^3 (6y^2 - 17xy + 7x^2)
= 2y^3(3y-7x)(2y-x)