Welcome, young mathematicians! Today, we are diving deep into Class 9 Maths Chapter 2, Exercise 2.2 (MP Board / NCERT). In this session, we will master the concept of the Value of a Polynomial and how to find the Zeroes of a Polynomial. Grab your notebooks, follow these step-by-step solutions, and let's make math simple and fun!
Exercise 2.2 Step-by-Step Solutions
Q1. Find the value of the polynomial 5x - 4x2 + 3 at:
Let p(x) = 5x - 4x2 + 3
-
(i) x = 0
p(0) = 5(0) - 4(0)2 + 3
p(0) = 0 - 0 + 3 = 3 -
(ii) x = -1
p(-1) = 5(-1) - 4(-1)2 + 3
p(-1) = -5 - 4(1) + 3
p(-1) = -5 - 4 + 3 = -9 + 3 = -6 -
(iii) x = 2
p(2) = 5(2) - 4(2)2 + 3
p(2) = 10 - 4(4) + 3
p(2) = 10 - 16 + 3 = -6 + 3 = -3
Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 - y + 1
- p(0) = (0)2 - 0 + 1 = 1
- p(1) = (1)2 - 1 + 1 = 1 - 1 + 1 = 1
- p(2) = (2)2 - 2 + 1 = 4 - 2 + 1 = 3
(ii) p(t) = 2 + t + 2t2 - t3
- p(0) = 2 + 0 + 2(0)2 - (0)3 = 2
- p(1) = 2 + 1 + 2(1)2 - (1)3 = 3 + 2 - 1 = 4
- p(2) = 2 + 2 + 2(2)2 - (2)3 = 4 + 2(4) - 8 = 4 + 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(0) = (0 - 1)(0 + 1) = (-1)(1) = -1
- p(1) = (1 - 1)(1 + 1) = (0)(2) = 0
- p(2) = (2 - 1)(2 + 1) = (1)(3) = 3
Q3. Verify whether the following are zeroes of the polynomial, indicated against them:
Note: If p(x) = 0 for a value, then that value is a zero of the polynomial.
- (i) p(x) = 3x + 1, x = -1/3:
p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0. Yes, it is a zero. - (ii) p(x) = 5x - π, x = 4/5:
p(4/5) = 5(4/5) - π = 4 - π ≠ 0. No, it is not a zero. - (iii) p(x) = x2 - 1, x = 1, -1:
p(1) = (1)2 - 1 = 0 and p(-1) = (-1)2 - 1 = 0. Yes, both are zeroes. - (iv) p(x) = (x + 1)(x - 2), x = -1, 2:
p(-1) = (-1 + 1)(-1 - 2) = 0(-3) = 0 and p(2) = (2 + 1)(2 - 2) = 3(0) = 0. Yes, both are zeroes. - (v) p(x) = x2, x = 0:
p(0) = 02 = 0. Yes, it is a zero. - (vi) p(x) = lx + m, x = -m/l:
p(-m/l) = l(-m/l) + m = -m + m = 0. Yes, it is a zero. - (vii) p(x) = 3x2 - 1, x = -1/√3, 2/√3:
p(-1/√3) = 3(-1/√3)2 - 1 = 3(1/3) - 1 = 1 - 1 = 0. (Yes for -1/√3)
p(2/√3) = 3(2/√3)2 - 1 = 3(4/3) - 1 = 4 - 1 = 3 ≠ 0. (No for 2/√3) - (viii) p(x) = 2x + 1, x = 1/2:
p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0. No, it is not a zero.
Q4. Find the zero of the polynomial in each of the following cases:
Tip: To find the zero of p(x), equate p(x) to 0 and solve for x.
| S.No. | Polynomial p(x) | Equation p(x) = 0 | Zero of the Polynomial |
|---|---|---|---|
| (i) | p(x) = x + 5 | x + 5 = 0 | x = -5 |
| (ii) | p(x) = x - 5 | x - 5 = 0 | x = 5 |
| (iii) | p(x) = 2x + 5 | 2x + 5 = 0 | x = -5/2 |
| (iv) | p(x) = 3x - 2 | 3x - 2 = 0 | x = 2/3 |
| (v) | p(x) = 3x | 3x = 0 | x = 0 |
| (vi) | p(x) = ax (a ≠ 0) | ax = 0 | x = 0 |
| (vii) | p(x) = cx + d (c ≠ 0, c, d are real numbers) | cx + d = 0 | x = -d/c |
ЁЯЪА Aptitude & NTSE Corner
Super Quick Mental Trick for NTSE & Olympiads:
To find the zero of any linear polynomial of the form ax + b, you do not need to solve it step-by-step. The zero is always x = -b/a (i.e., -Constant Term / Coefficient of x).
Example: For p(x) = 7x + 13, zero is directly -13/7. Save precious seconds in competitive exams!
We hope these detailed explanations for Class 9 Maths Chapter 2 Exercise 2.2 make your studies easier. Practice these problems yourself, and remember, consistency is the key to mastering algebra! If you have any questions, feel free to drop a comment below. Happy learning!
Academic Discussion & Doubts