**NCERT Solutions for Class 10 Maths Exercise 2.3 Chapter 2 Polynomials** which are provided here are prepared by our subject experts. These solutions are created and then reviewed by the NCERT subject experts to make it easy for the students to easily grasp the concept. This ensures that these NCERT Solutions are easily understandable by the Class 10 Maths students.

Division Algorithm for polynomials is explained in theÂ Solutions of Chapter 2 PolynomialsÂ Class 10 MathsÂ NCERT Exercise 2.3. These solutions are prepared keeping in mind that all NCERT guidelines shall be followed.Â It also ensures the whole syllabus is covered while providing the solutions, so that the student shall not miss out on any concept.

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**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}+5xâ€“3 , g(x) = x^{2}â€“2**

**Solution: **

Given,

Dividend = p(x) = x^{3}-3x^{2}+5xâ€“3

Divisor = g(x) = x^{2}â€“ 2

Therefore, upon division we get,

Quotient = xâ€“3

Remainder = 7xâ€“9

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

**Solution:**

Given,

Dividend = p(x) = x^{4 }â€“ 3x^{2 }+ 4x +5

Divisor = g(x) = x^{2} +1-x

Therefore, upon division we get,

Quotient = x^{2 }+ xâ€“3

Remainder = 8

**(iii) p(x) =x ^{4}â€“5x+6, g(x) = 2â€“x^{2}**

**Solution:**

Given,

Dividend = p(x) =x^{4} â€“ 5x + 6 = x^{4 }+0x^{2}â€“5x+6

Divisor = g(x) = 2â€“x^{2} = â€“x^{2}+2

Therefore, upon division we get,

Quotient = -x^{2}-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) t ^{2}-3, 2t^{4 }+3t^{3}-2t^{2}-9t-12 **

**Solutions: **

Given,

First polynomial = t^{2}-3

Second polynomial = 2t^{4 }+3t^{3}-2t^{2 }-9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t^{2}-3 is a factor of 2t^{4 }+3t^{3}-2t^{2 }-9t-12.

**(ii)x ^{2}+3x+1 , 3x^{4}+5x^{3}-7x^{2}+2x+2**

**Solutions: **

Given,

First polynomial = x^{2}+3x+1

Second polynomial = 3x^{4}+5x^{3}-7x^{2}+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x^{2} + 3x + 1 is a factor of 3x^{4}+5x^{3}-7x^{2}+2x+2.

**(iii) x ^{3}-3x+1, x^{5}-4x^{3}+x^{2}+3x+1**

**Solutions: **

Given,

First polynomial = x^{3}-3x+1

Second polynomial = x^{5}-4x^{3}+x^{2}+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, 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 âˆš(5/3) and â€“ âˆš(5/3).**

**Solutions: **

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

**âˆš(5/3) and â€“ âˆš(5/3) **are zeroes of polynomial f(x).

**âˆ´ **(x â€“**âˆš(5/3)**) (x+**âˆš(5/3) **= x^{2}-(5/3) = 0

**(3x ^{2}âˆ’5)=0,** is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x^{2}âˆ’5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x^{4Â }+6x^{3Â }âˆ’2x^{2Â }âˆ’10xâ€“5 = (3x^{2Â }â€“5)**(x ^{2}+2x+1)**

Now, on further factorizing (x^{2}+2x+1) we get,

**x ^{2}+2x+1**Â = x

^{2}+x+x+1 = 0

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

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

So, its zeroes are given by:Â **x= âˆ’1Â **and**Â x = âˆ’1.**

Therefore, all four zeroes of given polynomial equation are:

**âˆš(5/3),- âˆš(5/3) , âˆ’1 and âˆ’1.**

Hence, is the answer.

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

**by a polynomial g(x), the quotient and remainder were xâ€“2 and â€“2x+4, respectively. Find g(x).**

**Solution:**

Given,

Dividend, p(x) = x^{3}-3x^{2}+x+2

Quotient = x-2

Remainder = â€“2x+4

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

As we know,

Dividend = Divisor Ã— Quotient + Remainder

âˆ´ x^{3}-3x^{2}+x+2 = g(x)Ã—(x-2) + (-2x+4)

x^{3}-3x^{2}+x+2-(-2x+4) = g(x)Ã—(x-2)

Therefore, g(x) Ã— (x-2) = x^{3}-3x^{2}+3x-2

Now, for finding g(x) we will divide x^{3}-3x^{2}+3x-2 with (x-2)

Therefore, **g(x) = (x ^{2}â€“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)â‰ 0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor Ã— Quotient + Remainder

âˆ´ p(x) = g(x)Ã—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) = 3x^{2}+3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x^{2}+3x+3)/3 = x^{2}+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) = x^{2Â }+ 3 is a polynomial to be divided by g(x) = x â€“ 1.

So,Â x^{2Â }+ 3 = (x â€“ 1)Ã—(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) = x^{2Â }+ 1 is a polynomial to be divided by g(x) = x.

So, x^{2Â }+ 1 = (x)Ã—(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.

- Division Algorithm for Polynomials â€“ It includes five questions in which the first, second and fifth ones have three subparts each.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 2- Polynomials Exercise 2.3

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