Answer :

Given, p(1) = 0, p(√2) = 0 and p(– √2) = 0

Need to find the third degree polynomial p(x)

⇒ p(1) = 0 is given which satisfy p(x)

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

⇒ x – 1 is one factor of p(x)

⇒ p(√2) = 0 is given

∴ if x = √2 is substituted in p(x) then it satisfies the equation

⇒ x – √2 is another factor

⇒ p(– √2) = 0 is given

∴ if x = – √2 is substituted in p(x) then it satisfies the equation

⇒ x + √2 is third factor of the p(x)

⇒ Since, (x – 1)(x – √2)(x + √2) are the factors of the third degree polynomials

∴ p(x) = (x – 1)(x – √2)(x + √2)

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

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

⇒ p(x) = (x^{3} – x^{2} – 2x + 2)

Hence, x^{3} – x^{2} – 2x + 2 is the third degree polynomial which satisfies

p(1) = 0, p(√2) = 0 and p(– √2) = 0

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