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) = (x2 – x – √2x + √2)(x + √2)
⇒ p(x) = (x3 + √2x2 – x2 – √2x – √2x2 – 2x + √2x + 2)
⇒ p(x) = (x3 – x2 – 2x + 2)
Hence, x3 – x2 – 2x + 2 is the third degree polynomial which satisfies
p(1) = 0, p(√2) = 0 and p(– √2) = 0
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