Polynomials ⇒⇒ Exercise 2.2

Question 1

Find the value of the polynomial 5x - 4x² + 3
(i) x = 0
(ii) x = -1
(iii) x = 2

Solution :

Given polynomial 5x - 4x² + 3

Let P(x) = 5x - 4x² + 3

Now p(x) will mean replacing the variable with the value that is put in the bracket.

(i) putting x = 0

P(0) = 5(0) - 4(0)² + 3

P(0) = 0 - 0 + 3

P(0) = 3

Therefore, value of polynomial 5x - 4x² + 3 at x = 0 is 3

(ii) putting x = -1

P(-1) = 5(-1) - 4(-1)² + 3

P(-1) = -5 - 4 + 3

P(-1) = -6

Therefore, value of polynomial 5x - 4x² + 3 at x = -1 is -6

(iii) putting x = 2

P(2) = 5(2) - 4(2)² + 3

P(2) = 10 - 16 + 3

P(2) = -3

Therefore, value of polynomial 5x - 4x² + 3 at x = 2 is -3

Question 2 (i)

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² - y + 1

Solution :

Given polynomial p(y) = y² - y + 1

Now p(y) will mean replacing the variable with the value that is put in the bracket.

(i) putting putting y = 0

p(0) = (0)² - 0 + 1

p(0) = 0 - 0 + 1

p(0) = 1

Therefore, value of polynomial y² - y + 1 at y = 0 is 1

(ii) putting putting y = 1

p(1) = (1)² - 1 + 1

p(1) = 1 - 1 + 1

p(1) = 1

Therefore, value of polynomial y² - y + 1 at y = 1 is 1

(iii) putting y = 2

p(2) = (2)² - 2 + 1

p(2) = 4 - 2 + 1

p(2) = 3

Therefore, value of polynomial y² - y + 1 at y = 2 is 3

Question 2 (ii)

Find p(0), p(1) and p(2) for each of the following polynomials:
(ii) p(t) = 2 + t + 2t² - t³

Solution :

Given polynomial p(t) = 2 + t + 2t² - t³

Now p(t) will mean replacing the variable with the value that is put in the bracket.

(i) putting putting t = 0

p(0) = 2 + 0 + 2(0)² - (0)³

p(0) = 2 + 0 + 0 + 0

p(0) = 2

Therefore, value of polynomial 2 + t + 2t² - t³ at t = 0 is 2

(ii) putting putting t = 1

p(1) = 2 + 1 + 2(1)² - (1)³

p(1) = 2 + 1 + 2 - 1

p(1) = 4

Therefore, value of polynomial 2 + t + 2t² - t³ at t = 1 is 4

(iii) putting t = 2

p(2) = 2 + 2 + 2(2)² - (2)³

p(2) = 2 + 2 + 8 - 8

p(2) = 4

Therefore, value of polynomial 2 + t + 2t² - t³ at t = 2 is 4

Question 2 (iii)

Find p(0), p(1) and p(2) for each of the following polynomials:
(iii) p(x) = x³

Solution :

Given polynomial p(x) = x³

Now p(x) will mean replacing the variable with the value that is put in the bracket.

(i) putting putting x = 0

p(0) = (0)³

p(0) = 0

Therefore, value of polynomial x³ at x = 0 is 0

(ii) putting putting x = 1

p(1) = (1)³

p(1) = 1

Therefore, value of polynomial x³ at x = 1 is 1

(iii) putting x = 2

p(2) = (2)³

p(2) = 8

Therefore, value of polynomial x³ at x = 2 is 8

Question 2 (iv)

Find p(0), p(1) and p(2) for each of the following polynomials:
(iv) p(x) = (x – 1) (x + 1)

Solution :

Given polynomial p(x) = (x – 1) (x + 1)

Now p(x) will mean replacing the variable with the value that is put in the bracket.

(i) putting putting x = 0

p(0) = (0 – 1) (0 + 1)

p(0) = -1 × 1

p(0) = -1

Therefore, value of polynomial (x – 1) (x + 1) at x = 0 is -1

(ii) putting putting x = 1

p(1) = (1 – 1) (1 + 1)

p(1) = 0 × 2

p(1) = 0

Therefore, value of polynomial (x – 1) (x + 1) at x = 1 is 0

(iii) putting x = 2

p(2) = (2 – 1) (2 + 1)

p(2) = 1 × 3

p(2) = 3

Therefore, value of polynomial (x – 1) (x + 1) at x = 2 is 3

Question 3 (i)

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)= 3x + 1 , x = -1/3

Solution :

Given polynomial: p(x)= 3x + 1

When putting x = -1/3 , to check whether it is a zero.

Then,

P(${ -1 \over 3}$) = 3 × ($ { -1 \over 3}$) + 1

P(${ -1 \over 3}$) = -1 + 1

P(${ -1 \over 3}$) = 0

Therefore, x = -1/3 is the zero of given polynomial p(x).

Question 3 (ii)

Verify whether the following are zeroes of the polynomial, indicated against them.
(ii) p(x) = 5x – $ \pi $ , x = 4/ 5

Solution :

Given polynomial: p(x) = 5x – $ \pi $ ,

When putting x = 4/ 5 , to check whether it is a zero.

Then,

P(${ 4 \over 5}$) = 5 × ($ {4 \over 5}$) – $ \pi $

P(${ 4 \over 5}$) = 4 – $ \pi $

Since, (4 – $ {\pi} ) {\ne } $ 0

Therefore, x = ${ 4 \over 5}$ is not a zero of given polynomial p(x).

Question 3 (iii)

Verify whether the following are zeroes of the polynomial, indicated against them.
(iii) p(x)= x² -1 , x = 1, -1

Solution :

Given polynomial: p(x)= x² - 1

When putting x = 1 , to check whether it is a zero.

Then,

P(1) = (1)² - 1

P(1) = 1 - 1

P(1) = 0

(ii) Now , When putting x = -1 , to check whether it is a zero.

Then,

P(-1) = (-1)² - 1

P(-1) = 1 - 1

P(-1) = 0

Therefore, x = 1 and –1 both are zeroes of the given polynomial p(x).

Question 3 (iv)

Verify whether the following are zeroes of the polynomial, indicated against them.
(iv) p(x) = (x + 1) (x – 2) , x = -1, 2

Solution :

Given polynomial: p(x) = (x + 1) (x – 2)

When putting x = -1 , to check whether it is a zero.

Then,

P(-1) = (-1 + 1) (-1 – 2)

P(-1) = 0 × - 3

P(-1) = 0

(ii) Now , When putting x = 2 , to check whether it is a zero.

Then,

P(2) = (2 + 1) (2 – 2)

P(2) = 3 × 0

P(2) = 0

Therefore, x = -1 and 2 both are zeroes of the given polynomial p(x).

Question 3 (v)

Verify whether the following are zeroes of the polynomial, indicated against them.
(v) p(x)= x² , x = 0

Solution :

Given polynomial: p(x)= x² ,

When putting x = 0 , to check whether it is a zero.

Then,

P(0) = (0)²

P(0) = 0

Therefore, x = 0 is the zero of given polynomial p(x).

Question 3 (vi)

Verify whether the following are zeroes of the polynomial, indicated against them.
(vi) p(x) = lx + m , x = -m / l

Solution :

Given polynomial: p(x) = lx + m , ,

When putting x =${ -m \over l}$ , to check whether it is a zero.

Then,

P(${ -m \over l}$) = l × ${ -m \over l}$ + m

= -m + m

= 0

Therefore, x = ${ -m \over l}$ is the zero of given polynomial p(x).

Question 3 (vii)

Verify whether the following are zeroes of the polynomial, indicated against them.
(vii) p(x) = 3x² - 1 , x = ${ -1 \over {\sqrt 3}}$ , ${ 2 \over {\sqrt 3}}$

Solution :

Given polynomial: p(x)= 3x² - 1

(i) When putting x = ${ -1 \over {\sqrt 3}}$ , to check whether it is a zero.

Then,

P(${ -1 \over {\sqrt 3}}$) = 3 × $({ -1 \over {\sqrt 3}})^2$ - 1

= 3 × $({1 \over {3}})$ - 1

= 1 - 1

P(${ -1 \over {\sqrt 3}}$) = 0

(ii) Now , When putting x = ${ 2 \over {\sqrt 3}}$ , to check whether it is a zero.

Then,

P(${ 2 \over {\sqrt 3}}$) = 3 × $({ 2 \over {\sqrt 3}})^2$ - 1

= 3 × $({4 \over {3}})$ - 1

= 4 - 1

= 3

P(${ 2 \over {\sqrt 3}}$) = 3

Therefore, x = ${ -1 \over {\sqrt 3}}$ is the zero of given polynomial p(x) .

And, x = ${ 2 \over {\sqrt 3}}$ is not a zero of given polynomial p(x).

Question 3 (viii)

Verify whether the following are zeroes of the polynomial, indicated against them.
(viii) p(x) = 2x + 1 , x = ${1 \over 2}$

Solution :

Given polynomial: p(x)= 2x + 1

When putting x = ${1 \over 2}$ , to check whether it is a zero.

Then,

P(${1 \over 2}$) = 2 × (${1 \over 2}$)+ 1

= 1 + 1

= 2

P(${1 \over 2}$) = 2

Therefore, x = ${1 \over 2}$ is not a zero of given polynomial p(x).

Question 4 (i)

Find the zero of the polynomial in each of the following cases.
(i) p(x) = x + 5

Solution :

Given polynomial: p(x) = x + 5

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

x + 5 = 0

x = -5

Therefore, x = -5 is the zero of the given polynomial p(x).

Question 4 (ii)

Find the zero of the polynomial in each of the following cases.
(ii) p(x) = x - 5

Solution :

Given polynomial: p(x) = x - 5

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

x - 5 = 0

x = 5

Therefore, x = 5 is the zero of the given polynomial p(x).

Question 4 (iii)

Find the zero of the polynomial in each of the following cases.
(iii) p(x) = 2x + 5

Solution :

Given polynomial: p(x) = 2x + 5

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

2x + 5 = 0

2x = -5

x = ${-5 \over 2}$

Therefore, x = ${-5 \over 2}$ is the zero of the given polynomial p(x).

Question 4 (iv)

Find the zero of the polynomial in each of the following cases.
(iv) p(x) = 3x – 2

Solution :

Given polynomial: p(x) = 3x – 2

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

3x – 2 = 0

3x = 2

x = ${2 \over 3}$

Therefore, x = ${2 \over 3}$ is the zero of the given polynomial p(x).

Question 4 (v)

Find the zero of the polynomial in each of the following cases.
(v) p(x) = 3x

Solution :

Given polynomial: p(x) = 3x

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

3x = 0

x = ${0 \over 3}$

x = 0

Therefore, x = 0 is the zero of the given polynomial p(x).

Question 4 (vi)

Find the zero of the polynomial in each of the following cases.
(vi) p(x) = ax, a ≠0

Solution :

Given polynomial: p(x) = ax

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

ax = 0

x = ${0 \over a}$

x = 0

Therefore, x = 0 is the zero of the given polynomial p(x).

Question 4 (vii)

Find the zero of the polynomial in each of the following cases.
(vii) p(x) = cx + d, c ≠ 0, c, and d are real numbers

Solution :

Given polynomial: p(x) = cx + d

To find zero of polynomial, We put p(x) = 0 to and find x.

Thus,

p(x) = 0

cx + d = 0

cx = -d

x = ${-d \over c}$

Therefore, x = ${-d \over c}$ is the zero of the given polynomial p(x).