Find a seco

Given: p(1 + √3) = 0, p(1 – √3) = 0

Concept Used:

A second–degree polynomial is given by,

p(x) = ax² + bx + c

Explanation:

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) = x² – x + √3x –x + 1 – √3 – √3x + √3 – 3

⇒ p(x) = x² – 2x – 2

∴ x² – 2x – 2 is the second degree polynomial which satisfies p(1 + √3) = 0 and p(1 – √3) = 0.