NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.3

Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.3

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
Applying long division as follows

Therefore, we get
Quotient = x – 3 and Remainder = 7x – 9.

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x = x² – x +1
Applying long division as follows

Therefore, we get
Quotient = x² + x – 3 and Remainder = 8.

(iii) p(x) = x⁴ – 5x + 6 = x⁴ + 0x²– 5x + 6
g(x) = 2 – x² = – x² + 2
Applying long division as follows

Therefore, we get
Quotient = -x² – 2 and 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 + 3t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution :

(i) Applying long division as follows

Since we get, Remainder = 0, therefore the polynomial t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) Applying long division as follows

Since we get, Remainder = 0, therefore the polynomial x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) Applying long division as follows

Since we get, Remainder = 2, therefore the polynomial 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 and – .

Solution :

Given that and – are the two zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5.
Let p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Therefore, (x – ) and (x + ) are the factors of p(x).
Hence, (x² – ) is a factor of p(x).
For the other two factors applying long division as follows :

Thus,
3x⁴ + 6x³ – 2x² – 10x – 5
=  (x² – ) (3x² + 6x + 3)
= 3(x² – ) (x² + 2x + 1)
= 3(x² – ) (x² + x + x + 1)
= 3(x² – ) (x(x+1) + 1(x + 1))
= 3(x² – ) (x + 1)(x + 1)
Hence, the zeroes of the given polynomial are , –, -1, -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 :

Given
Divisor = g(x)
Quotient = x – 2
Dividend = x³ – 3x² + x + 2
Remainder = -2x + 4
We know that,
Dividend = Divisor x Quotient + Remainder
Therefore,
x³ – 3x² + x + 2 = g(x).(x – 2) + (-2x + 4)
g(x).(x – 2) = (x³ – 3x² + x + 2) – (-2x + 4)
g(x).(x – 2) = (x³ – 3x² + 3x – 2)
g(x) =
Applying long division as follows :

g(x) = x² – x + 1

5. Give examples of polynomials 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 p(x) = g(x)q(x) + r(x) , where q(x) , degree of r(x) = 0 or degree of r(x) < degree of g(x).

(i) deg p(x) = deg q(x)
p(x) = 2x² + 4x + 3
q(x) = x² + 2x + 1
g(x) = 2
r(x) = 1
the given condition is satisfied.

(ii) deg q(x) = deg r(x)
p(x) = x⁴ + x
q(x) = x
g(x) = x³
r(x) = x
the given condition is satisfied.

(iii) deg r(x) = 0
p(x) = 2x² + 4x + 3
q(x) = x² + 2x + 1
g(x) = 2
r(x) = 1
the given condition is satisfied.

Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.3

Click here for Maths Class 10 NCERT Solutions