Q. 24.2( 265 Votes )

Find p(0), p(1) a

Answer :

(i) p(y) = y2 – y + 1

p (0) = (0)2 – (0) + 1 = 1


p (1) = (1)2 – (1) + 1 = 1


p (2) = (2)2 – (2) + 1 = 3


(ii) p(t) = 2 + t + 2t2 – t3


p (0) = 2 + 0 + 2 (0)2 – (0)3 = 2


p (1) = 2 + (1) + 2 (1)2 – (1)3


= 2 + 1 + 2 – 1 = 4


p (2) = 2 + 2 + 2 (2)2 – (2)3


= 2 + 2 + 8 – 8 = 4


(iii) p(x) = x3


p (0) = (0)3 = 0


p (1) = (1)3 = 1


p (2) = (2)3 = 8


(iv) p(x) = (x – 1) (x + 1)


p (0) = (0 – 1) (0 + 1) = (-1) (1) = -1


p (1) = (1 – 1) (1 + 1) = 0 (2) = 0


p (2) = (2 – 1) (2 + 1) = 1 (3) = 3

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