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 + 2t2 - t3 iii) p(x) = x3 iv) p(x) = (x - 1)(x + 1)

Name: 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 + 2t2 - t3 iii) p(x) = x3 iv) p(x) = (x - 1)(x + 1)
Uploaded: 2020-04-30T14:09:18Z
Duration: 4 min 9 s
Description: The values of p(0), p(1), p(2) for each of the following polynomials p(y) = y2− y + 1, p(t) = 2 + t + 2t2 − t3, p(x) = x3, and p(x) = (x−1) (x+1) are {1, 1, 3}, {2, 4, 4}, {0, 1, 8}, and {-1, 0, 3} respectively.

The values of p(0), p(1), p(2) for each of the following polynomials p(y) = y2− y + 1, p(t) = 2 + t + 2t2 − t3, p(x) = x3, and p(x) = (x−1) (x+1) are {1, 1, 3}, {2, 4, 4}, {0, 1, 8}, and {-1, 0, 3} respectively.

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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 + 2t² - t³iii) p(x) = x³
iv) p(x) = (x - 1)(x + 1)

Solution:

To find the values p(0), p(1), and p (2) for each of the following polynomials we need to substitute the value of y as 0, 1, and 2 respectively in the given polynomial.

Let's look into the steps below.

(i) p(y) = y² - y + 1

p(0) = (0)² - (0) + 1 = 1

p(1) = (1)² - (1) + 1 = 1

p(2) = (2)² - 2 + 1 = 3

(ii) p(t) = 2 + t + 2t² - t³

p(0) = 2 + 0 + 2(0)² - (0)³

= 2 + 0 + 0 - 0 = 2

p(1) = 2 + 1 + 2(1)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) = x³

p(0) = (0)³ = 0

p(1) = (1)³ = 1

p(2) = (2)³ = 8

(iv) p(x) = (x - 1)(x + 1)

p(x) = x² - 1 [Using the identity (a + b)(a - b) = a² - b²]

p(0) = (0)² - 1 = -1

p(1) = (1)² - 1 = 0

p(2) = (2)² - 1 = 3

☛ Check: NCERT Solutions Class 9 Maths Chapter 2

Video Solution:

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 + 2t² - t³(iii) p(x) = x³(iv) p(x) = (x - 1)(x + 1)

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.2 Question 2

Summary:

The values of p(0), p(1), p(2) for each of the following polynomials p(y) = y²− y + 1, p(t) = 2 + t + 2t²− t³, p(x) = x³, and p(x) = (x−1) (x+1) are {1, 1, 3}, {2, 4, 4}, {0, 1, 8}, and {-1, 0, 3} respectively.

☛ Related Questions:

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 + 2t2 - t3 iii) p(x) = x3 iv) p(x) = (x - 1)(x + 1)

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 + 2t² - t³ (iii) p(x) = x³ (iv) p(x) = (x - 1)(x + 1)

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 + 2t² - t³iii) p(x) = x³
iv) p(x) = (x - 1)(x + 1)

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 + 2t² - t³(iii) p(x) = x³(iv) p(x) = (x - 1)(x + 1)