Chapter 2- Polynomials Exercise 2.3 – NCERT Solutions for class 10 Mathematics

Chapter 2- Polynomials Exercise 2.3 – NCERT Solutions for class 10

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

Solution:

Given,

Dividend = p(x) = x³-3x²+5x–3

Divisor = g(x) = x²– 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

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

Solution:

Given,

Dividend = p(x) = x⁴ – 3x² + 4x +5

Divisor = g(x) = x² +1-x

Therefore, upon division we get,

Quotient = x² + x–3

Remainder = 8

(iii) p(x) =x⁴–5x+6, g(x) = 2–x²

Solution:

Given,

Dividend = p(x) =x⁴ – 5x + 6 = x⁴ +0x²–5x+6

Divisor = g(x) = 2–x² = –x²+2

Therefore, upon division we get,

Quotient = -x²-2

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

Solutions:

Given,

First polynomial = t²-3

Second polynomial = 2t⁴ +3t³-2t² -9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t²-3 is a factor of 2t²+3t+4.

(ii)x²+3x+1 , 3x⁴+5x³-7x²+2x+2

Solutions:

Given,

First polynomial = x²+3x+1

Second polynomial = 3x⁴+5x³-7x²+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x² + 3x + 1 is a factor of 3x⁴+5x³-7x²+2x+2.

(iii) x³-3x+1, x⁵-4x³+x²+3x+1

Solutions:

Given,

First polynomial = x³-3x+1

Second polynomial = x⁵-4x³+x²+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, 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 √(5/3) and – √(5/3).

Solutions:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

∴ (x –√(5/3)) (x+√(5/3) = x²-(5/3) = 0

(3x²−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x²−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x⁴ +6x³ −2x² −10x–5 = (3x² –5)(x²+2x+1)

Now, on further factorizing (x²+2x+1) we get,

x²+2x+1 = x²+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer.

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,

Dividend, p(x) = x³-3x²+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor × Quotient + Remainder

∴ x³-3x²+x+2 = g(x)×(x-2) + (-2x+4)

x³-3x²+x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x³-3x²+x+2

Now, for finding g(x) we will divide x³-3x²+x+2 with (x-2)

Therefore, 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

Solutions:

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

(i) deg p(x) = deg q(x)

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, p(x) = 3x²+3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x²+3x+3)/3 = x²+x+1 = q(x)

Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

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

Let us take an example, p(x) = x² + 3 is a polynomial to be divided by g(x) = x – 1.

So, x² + 3 = (x – 1)×(x) + (x + 3)

Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

Hence, division algorithm is satisfied here.

(iii) deg r(x) = 0

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x² + 1 is a polynomial to be divided by g(x) = x.

So, x² + 1 = (x)×(x) + 1

Hence, quotient q(x) = x

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.