evaluate p(0) +p(1)+p(2) where p(t) = 2 + t + 2t^2 - t^3
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Answered by
3
(i) p(y)=y
2
−y+1
p(0)=(0)
2
−(0)+1=0+0+1=1
p(1)=(1)
2
−(1)+1=1−1+1=1
p(2)=(2)
2
−(2)+1=4−2+1=3
(ii) p(t)=2+t+2t
2
−t
3
p(0)=2+(0)+2(0)
2
−t
3
=2+0+0−0=2
p(1)=2+(1)+2(2)
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)=x
3
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
Answered by
2
Step-by-step explanation:
putting y=2
p(2)=(2)²-2+1
=4-2+1
=3
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