**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) = x ^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2**

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

**Solution :**

**(i) **p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2

Applying long division as follows

Therefore, we get

Quotient = x – 3 and Remainder = 7x – 9.

**(ii)** p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x = x^{2 }– x +1

Applying long division as follows

Therefore, we get

Quotient = x^{2} + x – 3 and Remainder = 8.

**(iii)** p(x) = x^{4} – 5x + 6 = x^{4} + 0x^{2}– 5x + 6

g(x) = 2 – x^{2} = – x^{2 }+ 2

Applying long division as follows

Therefore, we get

Quotient = -x^{2} – 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) t ^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12
(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2
(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1**

**Solution :**

**(i)** Applying long division as follows

Since we get, Remainder = 0, therefore the polynomial t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

**(ii)** Applying long division as follows

Since we get, Remainder = 0, therefore the polynomial x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

**(iii)** Applying long division as follows

Since we get, Remainder = 2, therefore the polynomial x^{3} – 3x + 1 is not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1.

**3. Obtain all other zeroes of 3x ^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are and – .**

**Solution :**

Given that and – are the two zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5.

Let p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

Therefore, (x – ) and (x + ) are the factors of p(x).

Hence, (x^{2} – ) is a factor of p(x).

For the other two factors applying long division as follows :

Thus,

3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

= (x^{2} – ) (3x^{2} + 6x + 3)

= 3(x^{2} – ) (x^{2} + 2x + 1)

= 3(x^{2} – ) (x^{2} + x + x + 1)

= 3(x^{2} – ) (x(x+1) + 1(x + 1))

= 3(x^{2} – ) (x + 1)(x + 1)

Hence, the zeroes of the given polynomial are , –, -1, -1.

**4. On dividing x ^{3} – 3x^{2} + 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 = x^{3} – 3x^{2} + x + 2

Remainder = -2x + 4

We know that,

Dividend = Divisor x Quotient + Remainder

Therefore,

x^{3} – 3x^{2} + x + 2 = g(x).(x – 2) + (-2x + 4)

g(x).(x – 2) = (x^{3} – 3x^{2} + x + 2) – (-2x + 4)

g(x).(x – 2) = (x^{3} – 3x^{2} + 3x – 2)

g(x) =

Applying long division as follows :

g(x) = x^{2} – 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) = 2x^{2} + 4x + 3

q(x) = x^{2} + 2x + 1

g(x) = 2

r(x) = 1

the given condition is satisfied.

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

p(x) = x^{4} + x

q(x) = x

g(x) = x^{3}r(x) = x

the given condition is satisfied.

**(iii)** deg r(x) = 0

p(x) = 2x^{2} + 4x + 3

q(x) = x^{2} + 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**

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