NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.2 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

**Question 1.
**

**Find the value of the polynomial 5x -4x**

^{2}+ 3 at**(i) x = 0**

**(ii) x = – 1**

**(iii) x = 2**

**Solution:**

Let p (x) = 5x – 4x

^{2}+ 3

**(i)**The value of p (x) = 5x – 4x

^{2}+ 3 at x= 0 is

p(0) = 5 x 0 – 4 x 0

^{2}+3

⇒ P (0) = 3

**(ii)**The value of p (x) = 5x – 4x

^{2}+ 3 at x = -1 is

p(-1) = 5(-D-4(-1)

^{2}+ 3 = – 5 -4 + 3

⇒ P(-1) = -6

**(iii)**The value of p (x) = 5x- 4x

^{2}+ 3 at x = 2 is

p (2) = 5 (2)- 4 (2)

^{2}+ 3= 10- 16+ 3

⇒ P (2) = – 3

**Question 2.
**

**Find p (0), p (1) and p (2) for each of the following polynomials.**

**(i) p(y) = y**

^{2}– y +1**(ii) p (t) = 2 +1 + 2t**

^{2}-t^{3 }(iii) P (x) = x^{3}**(iv) p (x) = (x-1) (x+1)**

**Solution:**

**(i)**p (y) = y

^{2}-y +1

∴ p(0) = 0

^{2}-0+1

⇒ p(0) = 1

p(1) = 1

^{2}-1+ 1

⇒ p(1) = 1

and p (2) = 2

^{2}– 2 + 1 =4-2+1

⇒ P (2) = 3

**(ii)**p (t) = 2 + t + 2t

^{2}-1

^{3 }p(0) = 2+ 0+ 2 x 0

^{2}– 0

^{3 }⇒ P (0) = 2

p (1) = 2 + 1 + 2 x 1

^{2}– 1

^{3 }⇒ p (1) = 3 + 2 – 1

⇒ p(1) = 4

and p (2) = 2 + 2 + 2 x 2

^{2}– 2

^{3 }=4+8-8

⇒ P (2) = 4

**(iii)**P(x) = x

^{3 }⇒ p (0) = 0

^{3}⇒ p (0) = 0 ⇒ p (1) = 1

^{3 }⇒ P (1) = 1

and p (2) = 2

^{3}⇒ p (2) = 8

**(iv)**p(x) = (x-1)(x+ 1)

p(0) = (0-1)(0+1)

⇒ P (0) = – 1

p (1) = (1 – 1) (1 + 1)

⇒ P (1) = 0

and p (2) = (2-1) (2+1)

⇒ P (2) = 3

**Question 3.
**

**Verify whether the following are zeroes of the polynomial, indicated against them.**

(i)p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\)

(i)p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\)

**(ii)p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\)**

**(iii) p (x) = x**

^{2}– 1, x = x – 1**(iv) p (x) = (x + 1) (x – 2), x = – 1,2**

**(v) p (x) = x**

^{2}, x = 0**(vi) p (x) = lx + m, x = – \(\frac { m }{ l }\)**

**(vii) P (x) = 3x**

^{2}– 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\)**(viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\)**

**Solution:**

**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 where c and d are real numbers.**

**Solution:**

**(i)**We have, p (x)= x+ 5

Now, p (x) = 0

⇒ x+ 5 = 0

⇒ x = -5

∴ – 5 is a zero of the polynomial p (x).

**(ii)**We have, p (x) = x – 5

Now, p (x) = 0

⇒ x – 5 = 0

⇒ x = 5

∴ 5 is a zero of the polynomial p (x).

**(iii)**We have, p (x) = 2x + 5

Now, P (x)= 0

⇒ 2x+ 5= 0

⇒ x = –\(\frac { 5 }{ 2 }\)

∴ –\(\frac { 5 }{ 2 }\) is a zero of the polynomial p (x).

**(iv)**We have, p (x)= 3x- 2

Now p(x) = 0

⇒ 3x- 2 = 0

⇒ x= \(\frac { 2 }{ 3 }\)

∴ \(\frac { 2 }{ 3 }\) is a zero of the polynomial p (x).

**(v)**We have, p (x) = 3x

Now, p (x)= 0

⇒ 3x=0

⇒ x =0

∴ 0 is a zero of the polynomial p (x).

**(vi)**We have, p (x)= ax, a ≠ 0

Now, p (x)= 0 ⇒ ax= 0

⇒ x= 0

∴ 0 is.a zero of the polynomial p (x).

**(vii)**We have, p (x) = cx + d,c ≠ 0

Now, p (x) = 0

⇒ cx + d = 0

x = – \(\frac { d }{ c }\)

∴ – \(\frac { d }{ c }\) is a zero of the polynomial p (x).

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