## EXERCISE 2.3

#### Page No 36:

#### Question 1:

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

(i)

(ii)

(iii)

#### Answer:

Quotient = *x* − 3

Remainder = 7*x* − 9

Quotient = *x*^{2} + *x* − 3

Remainder = 8

Quotient = −*x*^{2} − 2

Remainder = −5*x* +10

#### Question 2:

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

#### Answer:

**=**

Since the remainder is 0,

Hence, is a factor of .

Since the remainder is 0,

Hence, is a factor of .

Since the remainder ,

Hence, is not a factor of .

#### Question 3:

Obtain all other zeroes of , if two of its zeroes are .

#### Answer:

Since the two zeroes are ,

is a factor of** **.

Therefore, we divide the given polynomial by .

We factorize

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

*x* = −1

As it has the term , therefore, there will be 2 zeroes at* x* = −1.

Hence, the zeroes of the given polynomial are, −1 and −1.

#### Question 4:

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

#### Answer:

*g*(*x*) = ? (Divisor)

Quotient = (*x* − 2)

Remainder = (− 2*x* + 4)

Dividend = Divisor × Quotient + Remainder

*g*(*x*) is the quotient when we divide by

#### Question 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

#### Answer:

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 by 2.

Here*, p*(*x*) =

*g*(*x*) = 2

*q*(*x*) = 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)*

= 2()

=

Thus, the division algorithm is satisfied.

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

Let us assume the division of *x*^{3}* + x *by *x*^{2},

Here*, p*(*x*) = *x*^{3}* + x*

*g*(*x*) = *x*^{2}

*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)*

*x*^{3}* + x* = (*x*^{2 }) × *x* + *x*

*x*^{3}* + x = x*^{3}* + 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 *x*^{3}* + *1by *x*^{2}.

Here*, p*(*x*) = *x*^{3}* + *1

*g*(*x*) = *x*^{2}

*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 = (*x*^{2 }) × *x* + 1

*x*^{3}* + *1* = x*^{3}* + *1

Thus, the division algorithm is satisfied.