Question

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Question from ch 2 class 10​

Answer 1

Answer:

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The given polynomial is

p(x) = x⁴ - 6x³ - 26x² + 138x - 35

∵ (2 + √3) and (2 - √3) are the zeroes of p(x),

x - (2 + √3), x - (2 - √3) are factors of p(x)

⇒ {x - (2 + √3)} {x - (2 - √3)} is a factor of p(x)

⇒ x² - (2 - √3 + 2 + √3) x + (2 + √3) (2 - √3) is a factor of p(x)

⇒ x² - 4x + (4 - 3) is a factor of p(x)

⇒ x² - 4x + 1 is a factor of p(x)

∴ x⁴ - 6x³ - 26x² + 138x - 35

= x⁴ - 4x³ + x² - 2x³ + 8x² - 2x - 35x² + 140x - 35

= x² (x² - 4x + 1) - 2x (x² - 4x + 1) - 35 (x² - 4x + 1)

= (x² - 4x + 1) (x^2 - 2x - 35)

∴ another factor of p(x) is (x² - 2x - 35).

∴ x² - 2x - 35

= x² - 7x + 5x - 35

= x (x - 7) + 5 (x - 7)

= (x - 7) (x + 5)

∴ (x - 7), (x + 5) are two factors of p(x)

∴ the other two zeroes are 7, - 5.

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Answer 2

Answer:

The given polynomial is

p(x) = x⁴ - 6x³ - 26x² + 138x - 35

∵ (2 + √3) and (2 - √3) are the zeroes of p(x),

x - (2 + √3), x - (2 - √3) are factors of p(x)

⇒ {x - (2 + √3)} {x - (2 - √3)} is a factor of p(x)

⇒ x² - (2 - √3 + 2 + √3) x + (2 + √3) (2 - √3) is a factor of p(x)

⇒ x² - 4x + (4 - 3) is a factor of p(x)

⇒ x² - 4x + 1 is a factor of p(x)

∴ x⁴ - 6x³ - 26x² + 138x - 35

= x⁴ - 4x³ + x² - 2x³ + 8x² - 2x - 35x² + 140x - 35

= x² (x² - 4x + 1) - 2x (x² - 4x + 1) - 35 (x² - 4x + 1)

= (x² - 4x + 1) (x^2 - 2x - 35)

∴ another factor of p(x) is (x² - 2x - 35).

∴ x² - 2x - 35

= x² - 7x + 5x - 35

= x (x - 7) + 5 (x - 7)

= (x - 7) (x + 5)

∴ (x - 7), (x + 5) are two factors of p(x)

∴ the other two zeroes are 7, - 5.