y²(y + 1) = 8y + 12

          y³ + y² = 8y + 12

y³ + y² - 8y - 12 = 0

Candidates for rational zeros are ± the divisors of 12

±1, ±2, ±3, ±4, ±6, ±12

Try 1

 1|1 1 -8 -12
  |  1  2  -6 
   1 2 -6 -18

No, that doesn't give a remainder of 0,
so 1 is not a solution.

Try 2
 2|1 1 -8 -12
  |  2  6  -4 
   1 3 -2 -16
No, that doesn't give a remainder of 0,
so 2 is not a solution.

Try 3
 3|1 1 -8 -12
  |  3 12  12 
   1 4  4   0
Yes, that gives a remainder of 0,
so 3 is a solution.

And now we have factored the left
side of

    y³ + y² - 8y - 12 = 0   as
 (y - 3)(y² + 4y + 4) = 0

The quadratic factor will also factor

(y - 3)(y + 2)(y + 2) = 0
 
Use the zero factor property:

   y = 3, y = -2, y = -2

So 3 is a solution of multiplicity 1
and -2 is a solution of multiplicity 2.

Edwin