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

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 2 |

Exercise | Class 9 Chapter 2 Polynomials Exercise 2.2 |

Number of Questions Solved | 4 |

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

**NCERT TEXTBOOK EXERCISES**

**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 = –1/3****(ii)p (x) = 5x – π, x = 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 = – ml**

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

^{2}– 1, x = – 1/**√**3,2/**√**3**(viii) p (x) = 2x + 1, x = 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 = –5/2

∴ –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= 2/3

∴ 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 = – d/c

∴ – d/c is a zero of the polynomial p (x).

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 PDF

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