## Intermediate Algebra (6th Edition)

1. Rearrange expression to equal zero. $y^{3}$ = 9y $y^{3}$ - 9y = 0 2. Factor the expression. y($y^{2}$ - 9) y(y + 3)(y - 3) 3. Apply the zero factor property. a) y = 0 b) y + 3 = 0 c) y - 3 = 0 4. Solve each linear equation. a) y = 0 b) y = -3 c) y = 3 The solutions are {-3, 0, 3}.