NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Chapter 2 Polynomials Exercise 2.2

Question 1: Find the value of the polynomial 5x - 4x² + 3 at
i) x = 0
ii) x = -1
iii) x = 2
Answer:
i) x = 0
= 5x - 4x² + 3
= 5(0) - 4(0)² + 3
= 0 - 0 + 3
= 3

ii) x = -1
= 5x - 4x² + 3
= 5(-1) - 4(-1)² + 3
= -5 - 4 + 3
= -9 + 3
= -6

iii) x = 2
= 5x - 4x² + 3
= 5(2) - 4(2)² + 3
= 10 - 16 + 3
= -3

Question 2: Find p(0), p(1), 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)
Answer:
i) p(y) = y² - y + 1
For p(0),
= (0)² - 0 + 1
= 0 - 0 + 1
= 1
For p(1),
= (1)² - 1 + 1
= 1 - 1 + 1
= 1
For p(2),
= (2)² - 2 + 1
= 4 - 2 + 1
= 3

ii) p(t) = 2 + t + 2t² - t³
For p(0),
= 2 + 0 + 2(0)² - (0)³
= 2 + 0 + 0 - 0
= 2
For p(1),
= 2 + 1 + 2(1)² - (1)³
= 2 + 1 + 2 - 1
= 4
For p(2),
= 2 + 2 + 2(2)² - (2)³
= 2 + 2 + 8 - 8
= 4

iii) p(x) = x³
For p(0),
= (0)³
= 0
For p(1),
= (1)³
= 1
For p(2),
= (2)³
= 8

iv) p(x) = (x - 1)(x + 1)
For p(0),
= (0 - 1)(0 + 1)
= (0 × 0) + (0 × 1) - (1 × 0) - (1 × 1)
= 0 + 0 - 0 - 1
= -1
For p(1),
= (1 - 1)(1 + 1)
= (1 × 1) + (1 × 1) - (1 × 1) - (1 × 1)
= 1 + 1 - 1 - 1
= 0
For p(2),
= (2 - 1)(2 + 1)
= (2 × 2) + (2 × 1) - (1 × 2) - (1 × 1)
= 4 + 2 - 2 - 1
= 3

Question 3: 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
Answer:
i) p(x) = 3x + 1, x = -1/3
= 3(-1/3) + 1
= -1 + 1
= 0
Therefore, this is a zero of the polynomial.

ii) p(x) = 5x - π, x = 4/5
= 5(4/5) - π
= 4 - π
Therefore, this is not a zero of the polynomial.

iii) p(x) = x² - 1, x = 1, -1
For x = 1,
= (1)² - 1
= 1 - 1
= 0
Therefore, this is a zero of the polynomial.
For x = -1,
= (-1)² - 1
= 1 - 1
= 0
Therefore, this is a zero of the polynomial.

iv) p(x) = (x + 1)(x + 2), x = -1, 2
For x = -1
= (-1 + 1)(-1 + 2)
= (0)(1)
= 0
Therefore, this is a zero of the polynomial.
For x = 2
= (2 + 1)(2 + 2)
= (3)(4)
= 12
Therefore, this is not a zero of the polynomial.

v) p(x) = x², x = 0
= (0)²
= 0
Therefore, this is a zero of the polynomial.

vi) p(x) = lx + m, x = -m/l
= l(-m/l) + m
= -m + m
= 0
Therefore, this is a zero of the polynomial.

vii) p(x) = 3x² - 1, x = -1/√3, 2/√3
For x = -1/√3
= 3(-1/√3)² - 1
= 3(1/3) - 1
= 1 - 1
= 0
Therefore, this is a zero of the polynomial.

For x = 2/√3
= 3(2/√3)² - 1
= 3(4/3) - 1
= 4 - 1
= 3
Therefore, this is not a zero of the polynomial

viii) p(x) = 2x + 1, x = 1/2
= 2(1/2) + 1
= 1 + 1
= 2
Therefore, this is not a zero of the polynomial

Question 4: 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
Answer:
i) p(x) = x + 5
= x + 5 = 0
= x = 0 - 5
= x = -5
Therefore, -5 is the zero of the polynomial.

ii) p(x) = x - 5
= x - 5 = 0
= x = 0 + 5
= x = 5
Therefore, 5 is the zero of the polynomial.

iii) p(x) = 2x + 5
= 2x + 5 = 0
= 2x = 0 - 5
= 2x = -5
= x = -5/2
Therefore, -5/2 is the zero of the polynomial.

iv) p(x) = 3x - 2
= 3x - 2 = 0
= 3x = 0 + 2
= 3x = 2
= x = 2/3
Therefore, 2/3 is the zero of the polynomial.

v) p(x) = 3x
= 3x = 0
= x = 0/3
= x = 0
Therefore, 0 is the zero of the polynomial.

vi) p(x) = ax, a ≠ 0
= ax = 0
= x = 0/a
= x = 0
Therefore, 0 is the zero of the polynomial.

vii) p(x) = cx + d, c ≠ 0, c, d are real numbers
= cx + d = 0
= cx = 0 - d
= cx = -d
= x = -d/c
Therefore, -d/c is the zero of the polynomial.