Class 9 Chapter 2 Polynomials Exercise 2.2 Solutions

Polynomials Class 9 Exercise 2.2 teaches you how to evaluate polynomials by substituting values for variables. This is an important skill because it forms the basis of graphing, solving equations, and understanding polynomial behaviour.

These NCERT solutions explain each question clearly so that students can easily understand how to substitute values and calculate polynomial outputs. With proper practice, this exercise helps build accuracy and improves calculation speed.

Class 9 Chapter 2 Polynomials Exercise 2.2 Questions

The Class 9 Chapter 2 Polynomials Exercise 2.2 questions require you to find the value of a polynomial for different values of the variable. You will use substitution to calculate results for linear, quadratic, and cubic polynomials. Below we have provided NCERT Solutions Class 9 Chapter 2 Ex 2.2:

1. Find the value of the polynomial (x)=5x−4x 2 +3.

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution: Let f(x) = 5x−4x 2 +3 (i) When x = 0 f(0) = 5(0)-4(0) 2 +3 = 3 (ii) When x = -1 f(x) = 5x−4x 2 +3 f(−1) = 5(−1)−4(−1) 2 +3 = −5–4+3 = −6 (iii) When x = 2 f(x) = 5x−4x 2 +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)=y 2 −y+1

Solution: p(y) = y 2 –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+2t 2 −t 3

Solution: p(t) = 2+t+2t 2 −t 3 ∴ 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)=x 3

Solution: p(x) = x 3 ∴ 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) = x 2 −1, x = 1, −1

Solution: For, x = 1, −1; p(x) = x 2 −1 ∴ p(1)=1 2 −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) = x 2 , x = 0

Solution: For, x = 0 p(x) = x 2 p(0) = 0 2 = 0 ∴ 0 is a zero of p(x).

(vi) p(x) = lx +m, x = −m/ l

Solution: For, x = -m/ l ; p(x) = l x+m ∴ p(-m/ l) = l (-m/ l )+m = −m+m = 0 ∴ -m/ l is a zero of p(x).

(vii) p(x) = 3x 2 −1, x = -1/√3 , 2/√3

Solution: For, x = -1/√3 , 2/√3 ; p(x) = 3x 2 −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).

NCERT Maths Class 9 Polynomials Exercise 2.2 PDF

The NCERT Maths Class 9 Polynomials Exercise 2.2 PDF includes easy, step-by-step solutions that help students revise anytime. The PDF clearly shows how to substitute values correctly and compute results without confusion. 

The PDF is helpful as it includes simple solutions that do not skip steps. By practicing using this PDF, you can improve accuracy and avoid common mistakes.

NCERT Solutions Maths Class 9 PDF

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