1) Factor 4xReturn to the tutorial^{3}- 12x^{2}+ x - 3 SOLUTION: Factors of 4: 1, -1, 2, -2, 4, -4 (possible h values) Factors of -3: 1, -1, 3, -3 (possible k values) Taking only the positive h values, and all k values, the following are possible factors of 4x^{3}- 12x^{2}+ x - 3: (x + 1), (x - 1), (x + 3), (x - 3) (2x + 1), (2x - 1), (2x + 3), (2x - 3) (4x + 1), (4x - 1), (4x + 3), (4x - 3) Checking: (x + 1) factor: k = -1, substitute x = -1 into polynomial: 4(-1)^{3}- 12(-1)^{2}+ (-1) -3 = -20 Thus, not a factor (x - 1) factor: k = 1, sub in x = 1: 4(1)^{3}> - 12(1)^{2}+ (1) -3 = -10 Thus, not a factor (x + 3) factor: k = -3, sub in x = -3: 4(-3)^{3}> - 12(-3)^{2}+ (-3) -3 = -6 Thus, not a factor (x - 3) factor: k = 3, sub in x = 3: 4(3)^{3}> - 12(3)^{2}+ (3) -3 = 0 Thus, (x - 3) IS A FACTOR! Using long division to find other factor(s): 4x^{2}+ 0x + 1 ------------------- (x - 3)| 4x^{3}- 12x^{2}+ x - 3 4x^{3}- 12x^{2}---------- 0x^{2}+ x 0x^{2}- 0x --------- x - 3 x - 3 ----- 0 Therefore, (x - 3) and (4x^{2}+ 1) are the two factors of 4x^{3}- 12x^{2}+ x - 3