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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 âˆ’ 4x2 + 3

(i) x = 0

(ii) x = â€“1

(iii) x = 2

Solution:

Given equation: 5x âˆ’ 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) = 5xâˆ’4x2+3

f(âˆ’1) = 5(âˆ’1)âˆ’4(âˆ’1)2+3

= âˆ’5â€“4+3

= âˆ’6

(iii) When x = 2

f(x) = 5xâˆ’4x2+3

f(2) = 5(2)âˆ’4(2)2+3

= 10â€“16+3

= âˆ’3

### Question 2: Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y2âˆ’y+1

(ii) p(t) = 2+t+2t2âˆ’t3

(iii) p(x) = x3

(iv) P(x) = (xâˆ’1)(x+1)

Solution:

(i) p(y) = y2 â€“ y + 1

Given equation: p(y) = y2â€“y+1

Therefore, p(0) = (0)2âˆ’(0)+1 = 1

p(1) = (1)2â€“(1)+1 = 1

p(2) = (2)2â€“(2)+1 = 3

Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y2â€“y+1

(ii) p(t) = 2 + t + 2t2 âˆ’ t3

Given equation: p(t) = 2+t+2t2âˆ’t3

Therefore, p(0) = 2+0+2(0)2â€“(0)3 = 2

p(1) = 2+1+2(1)2â€“(1)3 = 2+1+2â€“1 = 4

p(2) = 2+2+2(2)2â€“(2)3 = 2+2+8â€“8 = 4

Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t2âˆ’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) = (xâˆ’1)(x+1)

Given equation: p(x) = (xâ€“1)(x+1)

Therefore, p(0) = (0â€“1)(0+1) = (âˆ’1)(1) = â€“1

p(1) = (1â€“1)(1+1) = 0(2) = 0

p(2) = (2â€“1)(2+1) = 1(3) = 3

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

### Question 3: Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x+1, x=âˆ’1/3

(ii) p(x) = 5xâ€“Ï€, x = 4/5

(iii) p(x) = x2âˆ’1, x=1, âˆ’1

(iv) p(x) = (x+1)(xâ€“2), x =âˆ’1, 2

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

(vi) p(x) = lx+m, x = âˆ’m/l

(vii) p(x) = 3x2âˆ’1, x = -1/âˆš3 , 2/âˆš3

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

Solution:

(i) p(x)=3x+1, x=âˆ’1/3

Given: p(x)=3x+1 and x=âˆ’1/3

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

For, x = -1/3

p(âˆ’1/3) = 3(-1/3)+1

= âˆ’1+1

= 0

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

(ii) p(x)=5xâ€“Ï€, x = 4/5

Given: p(x)=5xâ€“Ï€ 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)â€“Ï€

= 4â€“Ï€

Hence, p(x) of 4/5 â‰  0

(iii) p(x)=x2âˆ’1, x=1, âˆ’1

Given: p(x)=x2âˆ’1 and x=1, âˆ’1

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

For x = 1

p(1) = 12âˆ’1

=1âˆ’1

= 0

For, x = -1

p(âˆ’1) = (-1)2âˆ’1

= 1âˆ’1

= 0

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

(iv) p(x) = (x+1)(xâ€“2), x =âˆ’1, 2

Given: p(x) = (x+1)(xâ€“2) and x =âˆ’1, 2

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

For, x = âˆ’1

p(âˆ’1) = (âˆ’1+1)(âˆ’1â€“2)

= (0)(âˆ’3)

= 0

For, x = 2

p(2) = (2+1)(2â€“2)

= (3)(0)

= 0

Hence, p(x) of âˆ’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 = âˆ’m/l

Given: p(x) = lx+m and x = âˆ’m/l

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

For, x = âˆ’m/l

p(-m/l)= l(-m/l)+m

= âˆ’m+m

= 0

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

(vii) p(x) = 3x2âˆ’1, x = -1/âˆš3 , 2/âˆš3

Given: p(x) = 3x2âˆ’1 and x = -1/âˆš3 , 2/âˆš3

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

For, x = -1/âˆš3

p(-1/âˆš3) = 3(-1/âˆš3)2 -1

= 3(1/3)-1

= 1-1

= 0

For, x =  2/âˆš3

p(2/âˆš3) = 3(2/âˆš3)2 -1

= 3(4/3)-1

= 4âˆ’1

=3 â‰  0

Hence, p(x) of -1/âˆš3 = 0

but, p(x) of 2/âˆš3 â‰  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

= 2â‰ 0

Hence, p(x) of 1/2 â‰  0

### Question 4: Find the zero of the polynomials in each of the following cases:

(i) p(x) = x+5

(ii) p(x) = xâ€“5

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

(iv) p(x) = 3xâ€“2

(vii) p(x) = cx+d, c â‰  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 = âˆ’5

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

(ii) p(x) = xâ€“5

Given: p(x) = xâ€“5

p(x) = xâˆ’5

xâˆ’5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = xâ€“5 is when x = 5

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

Given: p(x) = 2x+5

p(x) = 2x+5

2x+5 = 0

2x = âˆ’5

x = -5/2

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

(iv) p(x) = 3xâ€“2

Given: p(x) = 3xâ€“2

p(x) = 3xâ€“2

3xâˆ’2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3xâ€“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, aâ‰  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 â‰  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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