NCERT Solutions Maths Class 10 Chapter 2

Go To:  Exercise 2.1    Exercise 2.2    Exercise 2.4

NCERT Solutions Maths Class 10 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

Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is x – 3 and the remainder is 7x – 9.

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

Note that g(x) is not in standard form. First write it in standard form and then divide.

Then, g(x) = x2 – x + 1

We stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is x2 + x – 3 and the remainder is 8.

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

Note that g(x) is not in standard form. First write it in standard form and then divide.

Then, g(x) = x2 + 2

We stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is –x2 – 2 and the remainder is −5+ 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) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12

Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Here, remainder = 0

Hence, the first polynomial is a factor of the second polynomial.

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

Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Here, remainder = 0

Hence, the first polynomial is a factor of the second polynomial.

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

Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Here, remainder ≠ 0

Hence, the first polynomial is not a factor of the second polynomial.

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

and .

Solution: Since two zeroes of (3x4 + 6x3 – 2x2 – 10x – 5) are

and  which means that

is a factor of (3x4 + 6x3 – 2x2 – 10x – 5).

Applying Division Algorithm to find more factors, we get

We have, p(x) = g(x) × q(x).

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

= (x2 – 5/3)3(x2 + 2x + 1)

= 3(x2 – 5/3)(x2 + x + x + 1)

= 3(x2 – 5/3)(x + 1)(x + 1)

Therefore, other two zeroes of (3x4 + 6x3 – 2x2 – 10x – 5) are −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:  Let p(x) = (x3 − 3x2 + x + 2), q(x) = (x – 2) and r(x) = (–2x + 4)

According to Division Algorithm of Polynomial, we have

p(x) = g(x).q(x) + r(x)

x3 − 3x2 + x + 2 g(x).(x − 2) + (−2x + 4)

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

x3 − 3x2 + 3x − 2 = g(x).(x − 2)

g(x) = (x3 − 3x2 + 3x – 2)/(x – 2)

So, dividing (x3 − 3x2 + 3x − 2) by (x − 2), we get

Therefore, we have
g(x) = (x3 − 3x2 + 3x – 2)/(x – 2) = 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:  (i) Let p(x) = 3x2 + 3x + 6, g(x) = 3

Dividing p(x) by g(x), we get

q(x) = x2 + x + 2

So, we can see in this example that deg p(x) = deg q(x) = 2

(ii) Let p(x) = x3 + 5 and g(x) = x2 – 1

Dividing p(x) by g(x), we get

q(x) = x

So, we can see in this example that deg q(x) = deg r(x) = 1

(iii) Let p(x) = x2 + 5x − 3, g(x) = x + 3

Dividing p(x) by g(x), we get

q(x) = x + 2

So, we can see in this example that deg r(x) = 0