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# Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5 (ii) p(x) = x - 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x - 2

(v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers

**Solution:**

In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0.

**(i)** p(x) = x + 5

p(x) = 0

⇒ x + 5 = 0

⇒ x = -5

Therefore, -5 is the zero of p(x).

**(ii)** p(x) = x - 5

p(x) = 0

⇒ x - 5 = 0

⇒ x = 5

Therefore, 5 is the zero of p(x).

**(iii)** p(x) = 2x + 5

p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = -5/2

Therefore, -5/2 is the zero of p(x).

**(iv**) p(x) = 3x - 2

p(x) = 0

⇒ 3x - 2 = 0

⇒ 3x = 2

⇒ x = 2/3

Therefore, 2/3 is the zero of p(x).

**(v)** p(x) = 3x

p(x) = 0

⇒ 3x = 0

⇒ x = 0

Therefore, 0 is the zero of p(x).

**(vi)** p(x) = ax, a ≠ 0

p(x) = 0

⇒ ax = 0

⇒ x = 0

Therefore, 0 is the zero of p(x).

**(vii)** p(x) = cx + d, c ≠ 0, c, d are real numbers.

p(x) = 0

⇒ cx + d = 0

⇒ cx = -d

⇒ x = -d/c

Therefore, -d/c is the zero of p(x).

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x - 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x - 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.2 Question 4

**Summary:**

The zeros of the polynomials in each of the following cases p(x) = x + 5, p(x) = x - 5, p(x) = 2x + 5, p(x) = 3x - 2, p(x) = 3x, p(x) = ax, (a ≠ 0), and p(x) = cx + d, (c ≠ 0 and c,d are real numbers) are -5, 5, -5/2, 2/3, 0, 0, and -d/c respectively.

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