Given, (ax + 2) (bx + 7) = 15x^2 + cx + 14 abx^2 + 7ax + 2bx + 14 = 15x^2 + cx + 14 abx^2 + (7a + 2b) x + 14 = 15x^2 + cx + 14 Comparing on both sides, ab = 15 (7a + 2b) = c And already given a + b = 8 We know that (a - b) ^2 = (a + b) ^2 - 4ab (a - b) ^2 = (8) ^2 - 4(15) (a - b) ^2 = 64 – 60 (a - b) ^2 = 4 (a - b) = √4 (a - b) = + 2 or (a - b) = - 2 Case 1: a - b = 2 a + b = 8→eq1 a - b = 2→eq2 Solving these two we get 2a = 10 a = 5 By substituting a value in eq1 5 + b = 8 b = 3 As c = (7a + 2b) c = (7(5) + 2(3)) c = 41. Case 2 :a - b = - 2 a + b = 8 a - b = - 2 Solving these two equations we get 2a = 6 a = 3 By substituting a value in eq1 3 + b = 8 b = 5 As c = (7a + 2b) c = (7(3) + 2(5)) c = 31. So the two possible values of c are 41 and 31.