# SOLUTION: solve y^2(y+1)=8y+12

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 Click here to see ALL problems on Polynomials-and-rational-expressions Question 638383: solve y^2(y+1)=8y+12Answer by Edwin McCravy(8921)   (Show Source): You can put this solution on YOUR website!``` 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```