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.
☛ Related Questions:
- Find the value of the polynomial 5x - 4x² + 3 ati) x = 0ii) x = -1iii) x = 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 + 2t2 - t3iii) p(x) = x3iv) p(x) = (x - 1)(x + 1)
- Verify whether the following are zeroes of the polynomial, indicated against them. (i) p(x) = 3x + 1, x = -(1/3)(ii) p(x) = 5x - π , x = 4/5(iii) p(x) = x2 - 1, x = 1, -1(iv) p(x) = (x + 1)(x - 2), x = -1, 2 (v) p(x) = x2 , x = 0(vi) p(x) = lx + m, x = -(m/l)(vii) p(x) = 3x2 - 1, x = -(1/√3), 2/√3(viii) p(x) = 2x + 1, x = 1/2
visual curriculum