F(x)=x^3-8x^2+29x-52 The possible rational zeros are ± the factors of 52, which are ±1,±2,±4,±13,±26, and ±52 You can keep trying those and the only one that gives a zero remainder and is therefore a rational zero is 4. 4|1 -8 29 -52 | 4 -16 52 1 -4 13 0 So we have factored F(x) this way: F(x) = (x-4)(x²-4x+13) To find the other zeros we set the factor x² - 4x + 13 = 0 and use the quadratic formula: x =x = x = x = x = x = x = x = 2 ± 3i So the other zeros are 2 + 3i and 2 - 3i So the factored form of F(x) is F(x) = (x - 4)[x - (2 + 3i)][x - (2 - 3i)] or F(x) = (x - 4)(x - 2 - 3i)(x - 2 + 3i) Edwin