Solution:- x - 1 = 0 x = 1 Let p(x) = Ax³ + Bx² - 36x + 22 p(1) = A(1)³ + B(1)² - (36 × 1) + 22 ⇒ A + B - 36 + 22 =0 ⇒ A + B - 14 = 0 ⇒ A + B = 14 ...........equation (1) 2^B = 64^A ⇒ 2^B = (2⁶)^A ⇒ B = 6A ⇒ A = B/6 Putting the value of A = B/6 in the equation, we get B/6 + B = 14 ⇒ B + 6B = 84 ⇒ 7B = 84 ⇒ B = 12 Putting the value of B = 12 in the equation A + B = 14, we get A + 12 = 14 ⇒ A =14 - 12 ⇒ A = 2 Hence A = 2 and B = 12 Hence, x - 1 is a factor of 2x³ + 12x² - 36x + 22 and if we divide it by x - 1, we will get 2x² + 14x - 22 as quotient and 0 as remainder. Let p(x) = 2x³ + 12x² - 36x + 22 p(1) = 2(1)³ + 12(1)² - (36 × 1) + 22 ⇒ 2 + 12 - 36 + 22 ⇒ 36 - 36 = 0