the value of p(y)=y^2-y+1is p=0
Answers
Answer:
Answer
Consider given the polynomial,
p(y)=y
2
−y+1 …….(1)
We have to find,p(0)=?
Put, y=0 in equation 1st ,
Now,
p(0)=0
2
−0+1
p(0)=1
Hence, this is the answer.(i) 1, 1 and 3 respectively.
(ii) 4, 5 and 6 respectively.
(iii)0, 1 and 2 respectively.
(iv) -1, 0 and 3 respectively.
Step-by-step explanation:
We need to find p(0), p(1) and p(2) for each of the following polynomials.
(i)
The given polynomial is
P(y)=y^2-y+1P(y)=y
2
−y+1
Substitute y=0,
P(0)=0^2-0+1=1P(0)=0
2
−0+1=1
Substitute y=1,
P(1)=1^2-1+1=1P(1)=1
2
−1+1=1
Substitute y=2,
P(2)=2^2-2+1=3P(2)=2
2
−2+1=3
Therefore, p(0), p(1) and p(2) are 1, 1 and 3 respectively.
Similarly
(ii)
The given polynomial is
P(t)=2+t+2P(t)=2+t+2
Find p(0), p(1) and p(2).
P(0)=2+0+2=4P(0)=2+0+2=4
P(1)=2+1+2=5P(1)=2+1+2=5
P(2)=2+2+2=6P(2)=2+2+2=6
Therefore, p(0), p(1) and p(2) are 4, 5 and 6 respectively.
(iii)
The given polynomial is
p(x)=xp(x)=x
Find p(0), p(1) and p(2).
P(0)=0P(0)=0
P(1)=1P(1)=1
P(2)=2P(2)=2
Therefore, p(0), p(1) and p(2) are 0, 1 and 2 respectively.
(iv)
The given polynomial is
P(x)=(x-1)(x+1)P(x)=(x−1)(x+1)
Find p(0), p(1) and p(2).
P(0)=(0-1)(0+1)=-1P(0)=(0−1)(0+1)=−1
P(1)=(1-1)(1+1)=0P(1)=(1−1)(1+1)=0
P(2)=(2-1)(2+1)=3P(2)=(2−1)(2+1)=3
Therefore, p(0), p(1) and p(2) are -1, 0 and 3 respectively.
#Learn more
Find p(0) p(1) for p( x )=x^2-x +1
https://brainly.in/question
Answer:
1
Step-by-step explanation:
Given => if the value of p is 0
So,
p(y) = y^2 - y + 1
p(0) = 0^2 - + 1
p(0) = 1
Hope you understand