Given: Our quadratic polynomial is ax2 + bx + c = 0 Concept used: If we substitute the obtained value when the divisor is equated with zero in the quadratic equation then it gives the remainder of the quadratic equation. Calculation: (x – 9) = 0, (x – 5) = 0 and (x + 10) = 0 Then, x = 9, x = 5 and x = –10 Putting different values of x in Quadratic equation Then, a(9)2 + 9b + c = 42 ⇒ 81a + 9b + c = 42 ----(1) And, a(5)2 + 5b + c = -34 ⇒ 25a + 5b + c = -34 ----(2) And, a(–10)2 – 10b + c = -34 ⇒ 100a – 10b + c = -34 ----(3) From eq (1) and eq (2) we get, 14a + b = 19 ----(5) From eq (2) and eq (3) we get, 5a = b ----(6) From eq (5) and eq (6) we get, a = 1 and b = 5 Now, a + b = 1 + 5 = 6 ∴ The value of (a + b) is 6