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Class 9 NCERT Solutions- Chapter 2 Polynomials – Exercise 2.2

  • Last Updated : 21 Dec, 2020

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

(i) x = 0

(ii) x = –1

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(iii) x = 2



Solution:

Given equation: 5x − 4x2 + 3

Therefore, 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

Question 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

Given equation: p(y) = y2–y+1

Therefore, p(0) = (0)2−(0)+1 = 1

p(1) = (1)2–(1)+1 = 1

p(2) = (2)2–(2)+1 = 3

Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y2–y+1

(ii) p(t) = 2 + t + 2t2 − t3

Given equation: p(t) = 2+t+2t2−t3

Therefore, 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

Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t2−t3

(iii) p(x) = x3

Given equation: p(x) = x3

Therefore, p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8



Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x3

(iv) p(x) = (x−1)(x+1)

Given equation: p(x) = (x–1)(x+1)

Therefore, 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

Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)

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) = 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

Given: p(x)=3x+1 and x=−1/3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/3

p(−1/3) = 3(-1/3)+1

= −1+1 

= 0

Hence, p(x) of -1/3 = 0

(ii) p(x)=5x–π, x = 4/5

Given: p(x)=5x–π and x = 4/5

Therefore, substituting the value of x in equation p(x), we get.

For, x = 4/5

p(4/5) = 5(4/5)–π 

= 4–π



Hence, p(x) of 4/5 ≠ 0

(iii) p(x)=x2−1, x=1, −1

Given: p(x)=x2−1 and x=1, −1

Therefore, substituting the value of x in equation p(x), we get.

For x = 1 

p(1) = 12−1

=1−1 

= 0

For, x = -1

p(−1) = (-1)2−1 

= 1−1 

= 0

Hence, p(x) of 1 and -1 = 0

(iv) p(x) = (x+1)(x–2), x =−1, 2

Given: p(x) = (x+1)(x–2) and x =−1, 2

Therefore, substituting the value of x in equation p(x), we get.

For, x = −1

p(−1) = (−1+1)(−1–2)

= (0)(−3) 

= 0

For, x = 2

p(2) = (2+1)(2–2) 

= (3)(0) 

= 0

Hence, p(x) of −1, 2 = 0

(v) p(x) = x2, x = 0

Given:  p(x) = x2 and x = 0

Therefore, substituting the value of x in equation p(x), we get.

For, x = 0

p(0) = 02 = 0

Hence, p(x) of 0 = 0

(vi) p(x) = lx+m, x = −m/l

Given: p(x) = lx+m and x = −m/l

Therefore, substituting the value of x in equation p(x), we get.

For, x = −m/l

p(-m/l)= l(-m/l)+m 

= −m+m 

= 0

Hence, p(x) of -m/l = 0

(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3

Given: p(x) = 3x2−1 and x = -1/√3 , 2/√3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/√3

p(-1/√3) = 3(-1/√3)2 -1 

= 3(1/3)-1 

= 1-1 

= 0

For, x =  2/√3

p(2/√3) = 3(2/√3)2 -1 

= 3(4/3)-1 



= 4−1

=3 ≠ 0

Hence, p(x) of -1/√3 = 0

but, p(x) of 2/√3 ≠ 0

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

Given: p(x) =2x+1 and x = 1/2

Therefore, substituting the value of x in equation p(x), we get.

For, x = 1/2

p(1/2) = 2(1/2)+1 

= 1+1 

= 2≠0

Hence, p(x) of 1/2 ≠ 0

Question 4: Find the zero of the polynomials 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  

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

Solution:

(i) p(x) = x+5  

Given: p(x) = x+5

To find the zero, let p(x) = 0

p(x) = x+5

0 = x+5

x = −5

Therefore, the zero of the polynomial p(x) = x+5 is when x = -5

(ii) p(x) = x–5

Given: p(x) = x–5

p(x) = x−5

x−5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = x–5 is when x = 5

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

Given: p(x) = 2x+5

p(x) = 2x+5

2x+5 = 0

2x = −5

x = -5/2

Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2

(iv) p(x) = 3x–2  

Given: p(x) = 3x–2  

p(x) = 3x–2

3x−2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3

(v) p(x) = 3x  

Given: p(x) = 3x  

p(x) = 3x

3x = 0

x = 0

Therefore, the zero of the polynomial p(x) = 3x   is when x = 0

(vi) p(x) = ax, a0

Given: p(x) = ax, a≠ 0

p(x) = ax

ax = 0

x = 0

Therefore, the zero of the polynomial p(x) = ax is when x = 0

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

Given: p(x) = cx+d

p(x) = cx + d

cx+d =0

x = -d/c

Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c




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