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

Quotient = x − 3

Remainder = 7x − 9

 

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

Quotient = x² + x − 3

Remainder = 8

 

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

Quotient = −x² − 2

Remainder = −5x +10

 

 

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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

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

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

Since the remainder ≠ 0

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

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

we divide the given polynomial by x² – ⁵/₃

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

= 3(x² – ⁵/₃)(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.

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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]

g(x) = x² – x + 1

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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.

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