Math, asked by smax86939, 3 months ago

the value of p(y)=y^2-y+1is p=0​

Answers

Answered by surajtilakdhari
0

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

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Answered by pushkardigraskar2005
0

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

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