Answer :

Given, p(1 + √3) = 0, p(1 – √3) = 0

Need to find a polynomial p(x) of second degree

⇒ since we know that p(1 + √3) = 0

∴ if x = 1 + √3 is substituted in p(x) then it satisfies the equation

⇒ x – (1 + √3) = 0 , and ((x – 1) – √3) is one factor of p(x)

And p(1 – √3) = 0 is given

⇒ if x = 1 – √3 is substituted in p(x) then it satisfies the equation

⇒ x – (1 – √3) = 0, and ((x – 1) + √3) is one factor of p(x)

⇒ since, ((x – 1) – √3) and ((x – 1) + √3) are the factors of p(x), it can be written as follows

⇒ p(x) = ((x – 1) – √3)((x – 1) + √3)

⇒ p(x) = (x – 1 – √3)(x – 1 + √3)

⇒ p(x) = x^{2} – x + √3x –x + 1 – √3 – √3x + √3 – 3

⇒ p(x) = x^{2} – 2x – 2

∴ x^{2} – 2x – 2 is the second degree polynomial which satisfies p(1 + √3) = 0 and p(1 – √3) = 0.

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