can you factor this problem completely?

12y^3 - 20y^2 + 30y - 50 = (12y - 20) * (y^2 + ay + b)

<=>

12y^3 - 20y^2 + 30y - 50 = 12y^3 + 12ay^2 + 12by - 20y^2 - 20ay - 20b

<=>

12y^3 - 20y^2 + 30y - 50 = 12y^3 + (12a - 20) y^2 + (12b - 20a) y - 20b


- 20 = 12a - 20 ===> a = 0

30 = 12b - 20a ===> b = 30/12 = 5/2

<=>

12y^3 - 20y^2 + 30y - 50 = (12y - 20) * (y^2 + 5/2)

<=>

12y^3 - 20y^2 + 30y - 50 = (6y - 10) * (2y^2 + 5)

<=>

12y^3 - 20y^2 + 30y - 50 = 2 * (3y - 5) * (2y^2 + 5)

*******

The first step is to divide everything by 2. That will make the expression a little simpler.
2(6y^3-10y^2+15y-25)
Now we're going to use the commutative property to just switch the two middle terms.
2(6y^3+15y-10y^2-25)
Now use the associative property to group the first two terms and the second two terms.
2[(6y^3+15y)+(-10y^2-25)]
The next step is to pull factors out of each small group of parenthesis. From the first one, we're going to pull out a 3y, and from the second we'll extract a -5
2[3y*(2y^2+5)+(-5)*(2y^2+5)]
Now each group in the brackets has the term (2y^2+5), so the next step is to apply the distributive property, but in reverse.
2[(3y-5)(2y^2+5)]
From here, we can eliminate the brackets
2*(3y-5)*(2y^2+5)
And there's your final answer. Unless you want to use imaginary numbers. In that case, you can further break it down to
2*(3y-5)*{y-[sqrt(2.5)*i]}*{y+[sqrt(2.5)*i]}
where sqrt(2.5) means the square root of 2.5. But as a standard, the answer is commonly accepted as

2*(3y-5)*(2y^2+5)

12y^3-20y^2+30y-50
=2(6y^3-10y^2+15y-25)
=2[2y^2(3y-5)+5(3y-5)]
=2(3y-5)(2y^2+5) answer//

(3y-5)(2y+sq.root10)(2y-sq.root10)

12y^3 - 20y^2 + 30y - 50
= 2(6y^3 - 10y^2 + 15y - 25)
= 2[(6y^3 - 10y^2) + (15y - 25)]
= 2[2y^2(3y - 5) + 5(3y - 5)
= 2(3y - 5)(2y^2 + 5)

12y^3-20y^2+30y-50
= 2y^2(6y-10)+5(6y-10)
= (2y^2+5)(6y-10)

2(3y - 5)(2y² -4y + 5)

same as the other one