_________________________ 2x - 1)ax³ - 2x² - 2x² + 3x - 5 We won't divide that out, but suppose we did. We would get some quotient Q(x) and a remainder of -8 Q(x) 2x - 1)ax³ - 2x² - 2x² + 3x - 5 ..... ...... _______ -8 Then use DIVIDEND = QUOTIENT*DIVISOR + REMAINDER ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8) We want to make the factor (2x - 1) equal to zero so that term will be eliminated. 2x - 1 = 0 2x = 1 x = ½ Now substitute x = ½ in ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8) a(½)³ - 2(½)² + 3(½) - 5 = Q(½)*(2*½ - 1) + (-8) a(⅛) - 2(¼) + - 5 = Q(½)*(1 - 1) - 8 - ½ + - 5 = Q(½)*(0) - 8 - ½ + - 5 = 0 - 8 - ½ + - 5 = -8 Clear of fractions by multiplying through by 8 a - 4 + 12 - 40 = -64 a - 32 = -64 a = -32 Checking: -16x² - 9x - 3 2x - 1)-32x³ - 2x² + 3x - 5 -32x³ + 16x² -18x² + 3x -18x² + 9x -6x - 5 -6x + 3 -8 It leaves the remainder -8, so we are correct! Edwin
