Answer :

Given, p(1) = 0, p(– 2) = 0

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

⇒ since we know that p(1) = 0

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

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

And p(– 2) = 0 is given

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

⇒ x + 2 = 0, and x + 2 is one factor of p(x)

⇒ since, x – 1 and x + 2 are the factors of p(x), it can be written as follows

⇒ p(x) = (x – 1)(x + 2)

⇒ p(x) = x^{2} – x + 2x – 2

⇒ p(x) = x^{2} + x – 2

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

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