Question 109731
12y^4 - 57y^3 + 36y^2
:
Although 57 may look prime, it is not. 3 * 19 = 57
:
We can factor out 3y^2:
3y^2(4y^2 - 19y + 12)
:
Then we can factor the quadratic to:
3y^2(4y - 3)(y - 4)