Question: Need to factor this, 6y^4-5y^3-5y^2+5y-1, equal to 0 and solve for y.
6y^4-5y^3-5y^2+5y-1 = 0
y = 1 is a root, so (y-1) is a factor.
(y-1)(6y^3 + y^2 - 4y +1) = 0
y = -1 is a root, so (y+1) is a factor.
(y-1)(y+1)(6y^2 - 5y + 1) = 0
The final quadratic factorises to give
(y-1)(y+1)(2y - 1)(3y - 1) = 0
The solutions are: y = 1, y = -1, y = 1/2, y = 1/3