Polynomials – Exercise 2.2 – Class IX

1. Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0

(ii) x = -1

(iii) x = 2

Solution:

P(x) = 5x – 4x² + 3

(i) x = 0 , P(0) = 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3

(ii) x = -1, P(-1) = 5(-1) – 4(-1)² + 3 = -5 – 4 + 3 = -9 + 3 = -6

(iii) x = 2, P(2)= 5(2)– 4(2)² + 3 = 10 – 16 + 3 = – 3

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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 – t³

(iii) p(x) = x³

(iv) p(x) = (x – 1)(x + 1)

Solution:

(i) p(y) = y² – y + 1

p(0) = 0² – 0 + 1 = 1

p(1) = 1² – 1 + 1 = 1

p(2) = 2² – 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 = 2 + 1 + 2 – 1 = 4

p(2) = 2 + (2) + 2(2)² – (2)³ = 2 + 2 + 8 – 8 = 4

 

(iii) p(x) = x³

p(0) = 0

p(1) = 1³ = 1

p(2) = 2³ = 8

 

(iv) p(x) = (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = -1 x 1 = -1

p(1) = (1 – 1)(1 + 1) = 0x2 = 0

p(2) = (2 – 1)(2 + 1) = 1 x 3 = 3

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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

Solution:

(i) p(x) = 3x + 1 , x = –¹/₃

Yes zeroes of the polynomial. 3x + 1 = 0 for x = –¹/₃

(ii) p(x) = 5x – π , x = ⁴/₅

No, 5x – π = 5(⁴/₅) – π= 4 – π ≠ 0 at x = ⁴/₅

(iii) p(x) = x² – 1 , x = 1, – 1

Yes. x² – 1 = 1² – 1 = 0 at x = 1,

x² – 1 = (-1)² – 1 = 1 – 1 = 0 at x = – 1

(iv) p(x) = (x + 1)(x – 2), x = -1, 2

yes.

(x + 1)(x – 2) = (-1 + 1)(-1 – 1) at  x = -1,

(x + 1)(x – 2) = (2 + 1)(2 – 2) = 3×0 = 0 at  x = 2

 

(v) p(x) = x² , x = 0

Yes. x² = 0² = 0

 

(vi) p(x) = lx + m , x = –^m/_l

Yes. l(-^m/_l) + m = -m + m = 0 at x = –^m/_l

 

(vii) p(x) = 3x² – 1 , x = – ¹/_√3, ²/_√3

3x² – 1 = 3(- ¹/_√3)² – 1 = 3(¹/₃) – 1 = 0  at x = – ¹/_√3

3x² – 1 = 3(- ²/_√3)² – 1 = 3(⁴/₃) – 1 = 4 – 1 = 3≠0  at –²/_√3

Therefore, at – ¹/_√3 is a zero but –²/_√3 is not a zero of the polynomial.

(viii) p(x) = 2x + 1, x = ¹/₂

2x + 1 = 2(¹/₂) + 1 = 2≠0 at x = ¹/₂

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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

Solution:

(i) p(x) = x + 5

x + 5 = 0 so, x = -5

Therefore, -5 is the zero of x + 5

(ii) p(x) = x – 5

x – 5 = 0 so, x = 5

Therefore, 5 is the zero of x – 5.

 

(iii) p(x) = 2x + 5

2x + 5 = 0 so, x = –⁵/₂

Therefore, –⁵/₂ is the zero of 2x + 5

 

(iv) p(x) = 3x – 2

3x – 2 = 0 so, x = ²/₃

Therefore, ²/₃ is the zero of 3x – 2

 

(v) p(x) = 3x

3x = 0 so, x = 0

Therefore, 0 is the zero of 3x

 

(vi) p(x) = ax , a ≠ 0

ax = 0 so, a = ⁰/_x

Therefore, ⁰/_x is the zero of ax

 

(vii) p(x) = cx + d , c≠0, c , d are real numbers

cx + d = 0 so, x = –^d/_c

Therefore, –^d/_c is the zero of cx + d

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