Polynomials – Exercise 2.3 – Class 10

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: 

(i) p(x) = x3 – 3x2 + 5x – 3 ; g(x) = x2 – 2

(ii) p(x) = x4 – 3x2 + 4x + 5 ; g(x) = x2 + 1 – x

(iii) p(x) = x4 – 5x + 6; g(x) = 2 – x2

Solution:

(i)p(x) = x3 – 3x2 + 5x – 3 ; g(x) = x2 – 2

2.png

Quotient = x − 3

Remainder = 7x − 9

 

(ii)p(x) = x4 – 3x2 + 4x + 5 ; g(x) = x2 + 1 – x

3.png

Quotient = x2 + x − 3

Remainder = 8

 

(iii) p(x) = x4 – 5x + 6; g(x) = 2 – x2

4

Quotient = −x2 − 2

Remainder = −5x +10

 

 


2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: 

(i) t2 – 3, 2t4 – 2t2 – 9t – 12

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Solution:

(i) t2 – 3, 2t4 – 2t2 – 9t – 12

t2 – 3 = t2 + 0.t – 3

5.png

Since the remainder is 0,

Hence, t2 – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12

 

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

6.png

Since the remainder is 0,

Hence x2 + 3x + 1 is a factor of 3x4 + 5x3 -7x2 + 2x + 2

 

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

7.png

Since the remainder ≠ 0

Hence x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1


3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x -5, if two of its zeroes are √(5/3) and -√(5/3)

Solution:

p(x) = 3x4 + 6x3 – 2x2 – 10x -5

Since the two zeros are √(5/3) and -√(5/3)

Therefore, [x – √(5/3)][x + √(5/3)] = (x25/3) is a factor of 3x4 + 6x3 – 2x2 – 10x -5. Thus

we divide the given polynomial by x25/3

8.png

3x4 + 6x3 – 2x2 – 10x -5 = (x25/3)(3x2 + 6x + 3)

= 3(x25/3)(x2 + 2x + 1)

We factorise x2 + 2x + 1 = (x + 1 )2

Therefore, its zero is given by x + 1 = 0 or x = −1

As it has the term (x+1)2 , therefore, there will be 2 zeroes at x = −1. Hence, the zeroes of the given polynomial are √(5/3) , – √(5/3), -1 and -1.


4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x − 2 and − 2x + 4, respectively. Find g(x).

Solution:

p(x) = x3 – 3x2 + x + 2 [dividend]

Quotient = (x -2)

Remainder = (-2x + 4)

Divisor = g(x)

Dividend = Divisor × Quotient + Remainder

x3 – 3x2 + x + 2 = g(x) x (x -2) + (-2x + 4)

x3 – 3x2 + x + 2 + 2x – 4 = g(x).(x – 2)

g(x) = [x3 – 3x2 + 3x – 2]/[x-2]

9.png

g(x) = x2 – x + 1


5. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x) 

(iii)deg r(x) = 0 

Solution:

According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0,

then we can find polynomials q(x) and r(x)

such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant). Let us assume the division of 6x2 + 2x + 2 by 2.

Here, p(x) = 6x2 + 2x + 2

g(x) = 2 q(x) = 3x2 + x + 1 and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

6x2 + 2x + 2 = (2)* (3×2 + x + 1) + 0

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of x3 + x by x2 ,

Here, p(x) = x3 + x g(x) = x2

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

x3 + x = (x2 ) × x + x

x3 + x = x3 + x

Thus, the division algorithm is satisfied.

(iii)deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x3 + 1 by x2 .

Here, p(x) = x3 + 1

g(x) = x2

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x 3 + 1 = (x2 ) × x + 1 x 3 + 1 = x3 + 1

Thus, the division algorithm is satisfied.


 

9 thoughts on “Polynomials – Exercise 2.3 – Class 10

Comments are closed.