x = 0; x = 2; x = 1+1; x = 1-i Get zero on the right, on the last three: x = 0; x-2 = 0; x-1-i = 0; x-1+i = 0 Multiply all four left sides together x(x-2)(x-1-i)(x-1+i) and set it equal to all four right sides multiplied together: (0)(0)(0)(0) = 0 x(x-2)(x-1-i)(x-1+i) = 0 (x²-2x)[(x-1)-i][(x-1+i] = 0 (x²-2x)[(x-1)²-i²] = 0 (x²-2x)[x²-2x+1-(-1)] = 0 (x²-2x)[x²-2x+1+1] = 0 (x²-2x)[x²-2x+2] = 0 x⁴-2x³+2x²-2x³+4x²-4x = 0 x⁴-4x³+6x²-4x = 0 P(x) = x⁴-4x³+6x²+4x²-4x Edwin