## 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.

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