Question

2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t)=2+t+2+ -
(iii) p(x)= x
(iv) p(x)=(x-1)(x+1)

please give answer to all questions. I will give my full points.​

Answer 1

(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

$P(y)=y^2-y+1$

Substitute y=0,

$P(0)=0^2-0+1=1$

Substitute y=1,

$P(1)=1^2-1+1=1$

Substitute y=2,

$P(2)=2^2-2+1=3$

Therefore, p(0), p(1) and p(2) are 1, 1 and 3 respectively.

Similarly

(ii)

The given polynomial is

$P(t)=2+t+2$

Find p(0), p(1) and p(2).

$P(0)=2+0+2=4$

$P(1)=2+1+2=5$

$P(2)=2+2+2=6$

Therefore, p(0), p(1) and p(2) are 4, 5 and 6 respectively.

(iii)

The given polynomial is

$p(x)=x$

Find p(0), p(1) and p(2).

$P(0)=0$

$P(1)=1$

$P(2)=2$

Therefore, p(0), p(1) and p(2) are 0, 1 and 2 respectively.

(iv)

The given polynomial is

$P(x)=(x-1)(x+1)$

Find p(0), p(1) and p(2).

$P(0)=(0-1)(0+1)=-1$

$P(1)=(1-1)(1+1)=0$

$P(2)=(2-1)(2+1)=3$

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/6236908

Answer 2

Answer:

here is your answer

For p (0),

p (y)=2y^2-3y+2

p(o) =2 (o)^2-3(o)+2

p(o)=2

Now for p (1),

p(y)=2y^2-3y+2

p(1)=2(1)^2-3(1)+2

p(1) =1

finally for p(2),

p(2)=2y^2-3y+2

p(2)=3