## Intermediate Algebra (6th Edition)

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