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