Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Question 1 Deleted for CBSE Board 2024 Exams

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Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 5 (Optional) Verify that 3, –1, (−1)/3 are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients. Let p(x) = 3x3 − 5x2 − 11x − 3 Verifying zeroes At x = 𝟑 p(𝟑) = 3 (3)^3 − 5 (3)^2 − 11 (3) − 3 = 3 (27) − 5 (9) − 33 − 3 = 81 − 45 − 33 − 3 = 81 − 81 = 0 Since p(3) = 0 ∴ 𝟑 is a zero of p(x) At x = −𝟏 p(−1) = 3 (−1)3 − 5 (−1)2 – 11 (−1) − 3 = 3 (−1) − 5 (1) + 11 − 3 = − 3 − 5 + 11 − 3 = − 11 + 11 = 0 Since p(−1) = 0 ∴ −1 is a zero of p(x) At x = – 𝟏/𝟑 p((−𝟏)/𝟑) = 3 ((−1)/3)^3 − 5((−1)/3)^2 − 11 ((−1)/3) − 3 = (−1)/9 − 5/9 + 11/3 − 3 = (−1 − 5 + 33 − 27)/9 = (−33+33)/9 = 0 Since p((−1)/3) = 0 ∴ ((−𝟏)/𝟑) is a zero of p(x). Verifying relationship between zeroes and coefficients For a cubic Polynomial p(x) = ax3 + bx2 + cx + d With zeroes α, 𝛽 and γ We have 𝛂 + 𝛽 + 𝛄 = (−𝒃)/𝒂 𝛂"𝛽" + 𝛽𝛄 + 𝛄𝛂 = 𝒄/𝒂 𝛂"𝛽" 𝛄= (−𝒅)/𝒂 For p(x) = 3x3 − 5x2 − 11x − 3 a = 3, b = − 5, c = − 11 and d = − 3 And zeroes are 𝜶 = 3, 𝜷 = − 1 and 𝜸 = (−1)/3 Now 𝜶+ 𝜷 + 𝜸 = 3 + (−1) +((−1)/3) = 3 − 1 − 1/3 = (9 − 3 − 1)/3 = 5/3 = (−𝒃)/𝒂 𝜶𝜷+ 𝜷𝜸 + 𝜸𝜶 = (3) (−1) + (−1) ((−1)/3) + ((−1)/3)(3) = −3 + 1/3 − 1 = (− 9 + 1 − 3)/3 = (−11)/3 = 𝒄/𝒂 𝜶𝜷𝜸 = 3 × (−1) ×((−1)/3) = 3/3 = 1 = (−𝒅)/𝒂 Hence, the relationship is verified