Question

how do i factor this out completely?
12y^5-34xy^4+14x^2y^3

Answer 1

You can first factor out 2y^3 and you get:

2y^3(6y^2-17xy+7x^2) Rearranging this slightly only because I like putting x first and because the middle term -17xy has x first:

2y^3(7x^2-17xy+6y^2)-- Now (7x^2-17xy+6y^2) can also be factored:

2y^3(7x-3y)(x-2y) This is the factor

Answer 2

\[2y ^{3}[(2y-x)(3y-7x)]\]