# NCERT solution class 10 chapter 2 polynomials exercise 2.3 mathematics

## EXERCISE 2.3

#### 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)

Quotient = x − 3

Remainder = 7x − 9

Quotient = x2 + x − 3

Remainder = 8

Quotient = −x2 − 2

Remainder = −5x +10

#### Question 2:

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

=

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 .

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 − 2 and − 2x + 4, respectively. Find g(x).

g(x) = ? (Divisor)

Quotient = (x − 2)

Remainder = (− 2x + 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

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 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 = (x) × 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 + 1by 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)

x3 + 1 = (x) × x + 1

x3 + 1 = x3 + 1

Thus, the division algorithm is satisfied.

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