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
