If one of the zeroes of the cubic polynomial  x 3 + ax 2 + bx + c is −1, then the product of the  other two zeroes is: 

(a)b – a + 1   (b) b – a – 1 

(c) a – b + 1  (d) a – b − 1

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Question 8 If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is −1, then the product of the other two zeroes is: b – a + 1 (b) b – a – 1 (c) a – b + 1 (d) a – b − 1 Let p(x) = x3 + ax2 + bx + c Given that one zero is −1 ∴ 𝜶 = −1, and we need to find product of other other two zeroes, i.e. 𝜷𝜸 We know that Product of Zeroes = (−𝐷)/𝐴 𝜶𝜷𝜸 = (−𝑐)/1 −1 × 𝛽𝛾 = −c 𝜷𝜸 = c Now, we need to c in terms a and b So, the correct answer is (A) 𝜶𝜷𝜸 = (−𝑐)/1 −1 × 𝛽𝛾 = −c 𝜷𝜸 = c Now, we need to c in terms a and b Since −1 is a zero of p(x) p(−1) = 0 Putting x = −1 (−1)3 + a(−1)2 + b(−1) + c = 0 −1 + a − b + c = 0 c = b − a + 1 From (1): Putting 𝜷𝜸 = c 𝜷𝜸 = b − a + 1 So, the correct answer is (A)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo