Class 9 Mathematics
Chapter 2: Polynomials
Exercise 2.2 – Stepwise Solutions
Important Formula Used
1. Value of polynomial: Substitute the given value of x in p(x)
2. Zero of polynomial: If p(a) = 0, then a is a zero of the polynomial
Q1. Find the value of polynomial 5x − 4x² + 3
Given: p(x) = 5x − 4x² + 3
(i) x = 0
p(0) = 5(0) − 4(0)² + 3 = 3
Answer: 3
(ii) x = −1
p(−1) = 5(−1) − 4(1) + 3
= −5 − 4 + 3 = −6
Answer: −6
(iii) x = 2
p(2) = 10 − 16 + 3 = −3
Answer: −3
Q2. Find p(0), p(1) and p(2)
(i) p(y) = y² − y + 1
p(0) = 1
p(1) = 1 − 1 + 1 = 1
p(2) = 4 − 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² − t³
p(0) = 2
p(1) = 2 + 1 + 2 − 1 = 4
p(2) = 2 + 2 + 8 − 8 = 4
(iii) p(x) = x³
p(0) = 0
p(1) = 1
p(2) = 8
(iv) p(x) = (x − 1)(x + 1)
= x² − 1
p(0) = −1
p(1) = 0
p(2) = 3
Q3. Verify zeroes
(i) p(x) = 3x + 1, x = −1/3
p(−1/3) = −1 + 1 = 0
Verified
(ii) p(x) = 5x − π, x = 4/5
p(4/5) = 4 − π ≠ 0
Not a zero
(iii) p(x) = x² − 1, x = 1, −1
p(1) = 0, p(−1) = 0
Both are zeroes
(iv) p(x) = (x + 1)(x − 2)
Zeroes = −1, 2 → Verified
(v) p(x) = x², x = 0
p(0) = 0
Verified
(vi) p(x) = lx + m, x = −m/l
p(−m/l) = −m + m = 0
Verified
(vii) p(x) = 3x² − 1
x = ±1/√3
p(1/√3) = 1 − 1 = 0
Verified
(viii) p(x) = 2x + 1, x = 1/2
p(1/2) = 1 + 1 = 2 ≠ 0
Not a zero
Q4. Find zero of polynomial
(i) p(x) = x + 5
x + 5 = 0 → x = −5
(ii) p(x) = x − 5
x = 5
(iii) p(x) = 2x + 5
2x = −5 → x = −5/2
(iv) p(x) = 3x − 2
x = 2/3
(v) p(x) = 3x
x = 0
(vi) p(x) = ax, a ≠ 0
x = 0
(vii) p(x) = cx + d
cx + d = 0 → x = −d/c
Final Answers Summary
- Q1 → 3, −6, −3
- Q2 → values calculated above
- Q3 → verified / not verified as shown
- Q4 → −5, 5, −5/2, 2/3, 0, 0, −d/c