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Question 2
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2y+1
(ii) p(t)=2+t+2t2t3
(iii) p(x)=x3
(iv) p(x)=(x1)(x+1)


Solution

(i) p(y)=y2y+1p(0)=020+1=1p(1)=121+1=1
p(2)=222+1=3. 

(ii) p(t)=2+t+2t2t3p(0)=2+0+2×0203=2p(1)=2+1+2×1213=4
p(2)=2+2+2×2223=4+88=4.

(iii) p(x)=x3p(0)=0p(1)=1
p(2)=8. 

(iv) p(x)=(x1)(x+1)p(0)=(1)(1)=1p(1)=(11)(1+1)=0
p(2)=(21)(2+1)=3 

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