Factor -4y-108y^4 completely

Question

-4y-108y^4=

Answer 1

The common factor is -4y.

-4y - 108y⁴ = -4y(1 + 27y³)

Note that (1 + 27y³) is a sum of cubes.

1 + 27y³ = (1 + 3y)[1 - 3y +(3y)²]

-4y - 108y⁴ = -4y(1 + 3y)[1 - 3y + (3y)²]

Answer 2

-4y-108y^4

-4y(1+27y^3)

the binomial is the sum of two cubes and is factorable; terms that are cubed are 1 and 3y

-4y(3y+1)(9y^2-3y+1)