If x + 1 is a factor of let p(x) = ax3 + x2 – 2x + 4a – 9 and g(x) = x + 1 Putting g(x) = 0 ⟹ x + 1= 0 ⟹ x = – 1 According to the factor theorem if g(x) is a factor of p(x) then p ( – 1) = 0 p( – 1) = a( – 1)3 + ( – 1)2 – 2( – 1) + 4a – 9 = 0 ⟹3a – 6 = 0 ⟹ a = 2