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Ex 2.4, 4 If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes. Let p(x) = x4 − 6x3 − 26x2 + 138 x − 35 Given roots are 2 ± √3 Since both are factors Since x = 2 + √𝟑 is a zero, x − (2 + √3) is a factor x − 2 – √3 is a factor Since x = 2 – √𝟑 is a zero, x − (2 – √3) is a factor x − 2 + √3 is a factor Hence, (x − 2 − √𝟑) × (x − 2 + √𝟑) is also a factor. ( (x − 2) − √3) × ( (x − 2) + √3) is also a factor. (x − 2)2 − (√3)2 x2 + 22 − 4x − 3 x2 + 4 − 4x − 3 x2 − 4x + 1 ∴ x2 − 4x + 1 is a factor of p(x) Now by dividing p(x) by (x2 − 4x + 1) We can find out other factors 𝑥2−4𝑥+ 1 − 𝟐𝒙 𝒙4 − 6𝒙𝟑 − 26𝒙^𝟐+ 138x − 35 Now, we find zeroes of x2 − 2x − 35 x2 – 2x − 35 = 0 We find the zeroes using Splitting the middle term method x2 − 7x + 5x − 35 = 0 x (x −7) + 5 (x − 7) = 0 (x + 5) (x − 7) = 0 So, x = –5 and x = 7 are zeros Therefore, the zeroes of p(x) are 2 + √𝟑 , 2 − √𝟑 , −5 and 7 Splitting the middle term method We need to find two numbers whose Sum = −2 & Product = −35

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.