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# Find p(0), p(1) and p (2) for each of the following polynomials:

(i) p(y) = y^{2} - y + 1 (ii) p(t) = 2 + t + 2t^{2} - t^{3}^{ }iii) p(x) = x^{3 }

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) = y^{2} - 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 + 2t^{2} - t^{3}

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) = x^{3}

p(0) = (0)^{3} = 0

p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8

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

p(x) = x^{2} - 1 [Using the identity (a + b)(a - b) = a^{2} - b^{2}]

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) = y^{2}− y + 1, p(t) = 2 + t + 2t^{2 }− t^{3}, p(x) = x^{3}, and p(x) = (x−1) (x+1) are {1, 1, 3}, {2, 4, 4}, {0, 1, 8}, and {-1, 0, 3} respectively.

**☛ Related Questions:**

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