find p(o),p(1) for each
(a) p(x)=4x^2+x-5
(b) p(t)=(t+1)(t-1)
Answers
Answered by
1
Answer:
a. Case-1) P(0) = 4(0)²+0-5
= 0+0-5
=-5.
Case-2) P(1) = 4(1)²+1-5
= 5-5
= 0
b. Case-1) P(0)= (0+1)(0-1)
= 1×(-1)
=-1
Case-2) P(1)= (1+1)(1-1)
2×0
= 0.
Answered by
1
Answer:
Step-by-step explanation:
(a) p(0)=4×0^2+0-5
=0-5
=-5
p(1)=4×1^2+1-5
=4+1-5
=5-5
=0
(b) p(0)=(0+1)(0-1)
=1×-1
=-1
p(1)=(1+1)(1-1)
=2×0
=0
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