NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials Exercise 2.2

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 provide a detailed and step-wise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. These solutions help the students memorising the method in which the different types of questions in this exercise are solved. The NCERT Class 9 Maths Solutions aim to improve the problem solving skills of students.

The concepts discussed in Class 9 are of priority as it would be continued even in higher education. The NCERT Solutions are always prepared by following NCERT guidelines so that they should cover the whole syllabus accordingly. These are very helpful in scoring well in board examinations.

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.2

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Access Answers of Maths NCERT Class 9 Chapter 2 – Polynomials Exercise 2.2

1. Find the value of the polynomial (x)=5x−4x²+3 

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

Let f(x) = 5x−4x²+3

(i) When x = 0

f(0) = 5(0)-4(0)²+3

= 3

(ii) When x = -1

f(x) = 5x−4x²+3

f(−1) = 5(−1)−4(−1)²+3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x²+3

f(2) = 5(2)−4(2)²+3

= 10–16+3

= −3

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

(i) p(y)=y²−y+1

Solution:

p(y) = y²–y+1

∴p(0) = (0)²−(0)+1=1

p(1) = (1)²–(1)+1=1

p(2) = (2)²–(2)+1=3

(ii) p(t)=2+t+2t²−t³

Solution:

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

Solution:

p(x) = x³

∴p(0) = (0)³ = 0

p(1) = (1)³ = 1

p(2) = (2)³ = 8

(iv) P(x) = (x−1)(x+1)

Solution:

p(x) = (x–1)(x+1)

∴p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

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

(i) p(x)=3x+1, x=−1/3

Solution:

For, x = -1/3, p(x) = 3x+1

∴p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x)=5x–π, x = 4/5

Solution:

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- π= 4-π

∴ 4/5 is not a zero of p(x).

(iii) p(x)=x²−1, x=1, −1

Solution:

For, x = 1, −1;

p(x) = x²−1

∴p(1)=1²−1=1−1 = 0

p(−1)=(-1)²−1 = 1−1 = 0

∴1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

Solution:

For, x = −1,2;

p(x) = (x+1)(x–2)

∴p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴−1,2 are zeros of p(x).

(v) p(x) = x², x = 0

Solution:

For, x = 0 p(x) = x²

p(0) = 0² = 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l

Solution:

For, x = -m/l ; p(x) = lx+m

∴ p(-m/l)= l(-m/l)+m = −m+m = 0

∴-m/l is a zero of p(x).

(vii) p(x) = 3x²−1, x = -1/√3 , 2/√3

Solution:

For, x = -1/√3 , 2/√3 ; p(x) = 3x²−1

∴p(-1/√3) = 3(-1/√3)²-1 = 3(1/3)-1 = 1-1 = 0

∴p(2/√3 ) = 3(2/√3)²-1 = 3(4/3)-1 = 4−1=3 ≠ 0

∴-1/√3 is a zero of p(x) but 2/√3  is not a zero of p(x).

(viii) p(x) =2x+1, x = 1/2

Solution:

For, x = 1/2 p(x) = 2x+1

∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0

∴1/2 is not a zero of p(x).

4. Find the zero of the polynomials in each of the following cases.

(i) p(x) = x+5 

Solution:

p(x) = x+5

⇒ x+5 = 0

⇒ x = −5

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

(ii) p(x) = x–5

Solution:

p(x) = x−5

⇒ x−5 = 0

⇒ x = 5

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

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

Solution:

p(x) = 2x+5

⇒ 2x+5 = 0

⇒ 2x = −5

⇒ x = -5/2

∴x = -5/2 is a zero polynomial of the polynomial p(x).

(iv) p(x) = 3x–2 

Solution:

p(x) = 3x–2

⇒ 3x−2 = 0

⇒ 3x = 2

⇒x = 2/3

∴x = 2/3  is a zero polynomial of the polynomial p(x).

(v) p(x) = 3x 

Solution:

p(x) = 3x

⇒ 3x = 0

⇒ x = 0

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

(vi) p(x) = ax, a≠0

Solution:

p(x) = ax

⇒ ax = 0

⇒ x = 0

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

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

Solution:

p(x) = cx + d

⇒ cx+d =0

⇒ x = -d/c

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

Access other exercise solutions of Class 9 Maths Chapter 2 – Polynomials

Students can access the NCERT Solutions of Class 9 Maths Chapter 2 using the links below.

Exercise 2.1 Solutions 5 Questions
Exercise 2.3 Solutions 3 Questions
Exercise 2.4 Solutions 5 Questions
Exercise 2.5 Solutions 16 Questions