NCERT Solutions For Class 10 Maths Chapter 2

NCERT Solutions Class 10 Maths Polynomials

NCERT Solutions for class 10 maths chapter 2 Polynomials, available on our website at BYJU’S includes solutions for all the questions present in class 10 Maths NCERT textbooks. These solutions are prepared by the subject experts to help students in their academics.

NCERT Solutions for class 10 Maths chapter 2 pdf includes step by step detail explanations for all the problems, including indeterminate, coefficients, non-negative integers, the degree of a polynomial and lot more. Students preparing for class 10 CBSE board exams can solve more problems from these NCERT solutions for class 10 maths chapter 2 as it contains more important questions for the topic Polynomials.

NCERT Solutions Class 10 Maths Chapter 2 Exercises

DEGREE OF A POLYNOMIAL:-

If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of polynomial p(x).

And expressions like:- x2+2,1x1etc., are not polynomials.

 

Example-1:

3x4+2x2+6x = 1, the highest power of the given polynomial equation is 4

Therefore, its degree is 4.

 

Example-2:

x5y3z + 2xy3+4x2yz2

This equation is of multiple variables (x, y, z) and to find the degree of this equation, we just need to add up the degrees of the variables in each of the terms and also it does not matter whether they are different variables.

Therefore, 1st term (5+3+1), 2nd term (1+3) and 3rd term (2+1+2).

The highest total is 9, of the 1st term,

Therefore, its degree is 9.

Polynomial equation with degree 1 is called a linear polynomial.

Example:     x – 3 = 0

Polynomial equation with degree 2 is called a quadratic polynomial.

Example:    x2+2x+7=0.

Polynomial equation with degree 3 is called a cubic polynomial.

Example:    3x35x2+7x+5=0.

 

Factorization of a quadratic polynomial equation by splitting middle term:-

General form of quadratic polynomial equation:- ax2 + bx + c = 0

Step 1:- Find the product of a (first term) and c (last term).

Step 2:- Then split the product of a and c into two numbers such that their product remains same as (a × c) and their sum or difference is equal to the value of b (middle term).

Step 3:- Now, there will be total four terms in the equation, make two groups with each group having two terms each. Then take whatever is common and equate it to 0, the resultant values of x will be factors of the quadratic equation.

 

Example:

12 x2 = -11x +15

Solution:

12x 2 + 11x -15 = 0                (product of a and c is 12×15=180)

12x 2 + 20x – 9x -15 = 0      (splitting 180 into 20 and 9 such that, 20 – 9 = 11 and 20×9 = 180).

4x(3x + 5) – 3(3x + 5) = 0

(4x -3)(3x + 5) = 0

4x – 3 = 0 or 3x + 5 = 0

4x = 3 or 3x = – 5

x = 34 or x = 53

Therefore, solution of this quadratic equation  is (53, 34 ).

This solution is called zeroes or roots of the given polynomial.

 

                                                                          EXERCISE 2.1

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

Example 1:-

EX 2.1 EP1

In the above graph, the curve intersects x axis at A and B.

Therefore, the equation of the curve will have maximum of 2 zeroes.

 

Example 2:-

2

From the above graph, the curve cuts the x-axis at point A

Therefore, number of zeroes of p(x) = 1

 

 

Example 3: –

From the above graph, the curve p(x) intersects x-axis at points A, B and C.

Therefore, the number of zeroes of polynomial equation is 3.

3

 

 

Zeroes and Coefficient of a polynomial: 

If α and β are zeroes of any quadratic polynomial ax2 + bx + c=0,

Then, Sum of zeroes = α+β = ba

Product of zeroes = αβ = ca

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: 

x2 – (α + β)x + αβ = 0                                    

   

                                    EXERCISE 2.2

Q.1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) 6x² + 11x + 5 = 0

Soln: –

6x2 + 11x + 5 = 6x2 + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x2 +11x +5 are { −1, 56 }

Now, Sum of zeroes of this given polynomial equation = −1+( 56 ) = 116

But, the Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= 116

And Product of these zeroes will be = 1×56 = 56

But, the Product of zeroes of any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 56

Hence the relationship is verified.

 

(ii) 4s2 – 4s + 1

Sol:

     4s2 – 4s + 1 = 4s2 – 2s 2s + 1

= 2s (2s 1) 1(2s 1)

= (2s 1) (2s 1)

∴ zeroes of the given polynomial are: {12,12}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofscoeff.ofs2

= 44 = 1

And the Product of these zeroes will be = 12×12
=14

But, Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofs2

= 14.

Hence, the relationship is verified.

 

(iii) 6x2 – 3 – 7x

Sol:

     6x2 – 7x – 3 = 6x2 – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (13,32)

∴ sum of these zeroes will be =  13+32
=76

But, The Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= 76

And Product of these zeroes will be = 13×32=12
Also, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 36 = 12

Hence, the relationship is verified.

 

(iv) 4u2 + 8u

Sol:

     4u2 + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = coeff.ofucoeff.ofu2

 = 84 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofu2                                                                          = 04 = 0

Hence, the relationship is verified.

 

(v) t2 – 15

Sol:

     t2 – 15 = (t+ 15) (t − 15)

Therefore, zeroes of the given polynomial are: – {15, −15}

∴ sum of these zeroes will be = 15 −  15 = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

                                                                = 01 = 0   

And the product of these zeroes will be = (15) × (−15)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constanttermcoeff.oft2                                                                     = 151  = −15

Hence, the relationship is verified.

 

(vi) 3x2 – x – 4

Sol:

 3x2 − x − 4   = 3x2 – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 43 }

∴ sum of these zeroes will be = −1 +43 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= (1)3 = 13

And the Product of these zeroes will be = {−1 × 43 }

= 43

But, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 43

Hence, the relationship is verified.

 

Q2. Form a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i). 26 , −3

Sol.

Given,

α + β = 26

αβ = −3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

Thus, the required quadratic equation will be:

x2 – (26)x −3 = 0

6x2 − 2x – 18 = 0.

 

(ii). 3 , 43

Sol.

Given,

α + β = 3

αβ = 43

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (3)x + 43 =0

3x2 33x + 4 = 0.

 

(iii).  0, 7

Sol.

Given,

α + β = 0

αβ = 7

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (0)x + 7 = 0

x2 + 7 = 0.

 

(iv).  −2, −2

Sol.

Given,

α + β = −2

αβ = −2

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α+β)x +αβ=0

∴ The required quadratic polynomial will be:

x2 – (−2)x −2 = 0

x2 + 2x – 2 = 0.

 

(v). 72, 39

Sol.

Given,

α + β = 72

αβ = 39

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:

x2 – (72)x + 39 = 0

18x2 + 63x + 6 = 0.

 

(vi). 6, 0

Sol.

Given,

α + β = 6

αβ = 0

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:-

x2 – 6x + 0 = 0

x2 – 6x = 0.

 

                                  EXTRA QUESTIONS

Q.1 Find a quadratic polynomial whose zeroes are: –2+12, 212.

Sol.

Given: –

α + β = 2+12 + 212

= 4.

αβ = (2+12)(212)=412=72

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

Thus, the required quadratic polynomial equation will be :-

x2 – (4)x −72 = 0

2x2−8x+7 = 0.

 

Q.2 If α and β are the roots of a quadratic polynomial ax2+bx+c, then find the value of  α2 + β2.

Sol.

From the equation, (α+β=ba)

And,               α×β=ca

(α+β)2=α2+β2+(2αβ)

α2+β2=(α+β)22αβ

α2+β2=(ba)22ca

α2+β2=b2a22ca

 ∴α2+β2=b22aca2.

Similarly, we can find out the values of (α3 + β3) and (α3 – β3).

 

Q3. If α and β are zeroes of a quadratic polynomial x2+4x+3, form the polynomial whose zeroes are  1+αβand1+βα.

Sol.

Since α and β are zeroes of a quadratic polynomial x2+4x+3,

α+β= -4, αβ = 3

Given: –  α1 = 1+αβ

β1 = 1βα

Now, sum of zeroes = 1+αβ+1+βα

= 2+αβ+βα

= 2αβ+α2+β2αβ

= (α+β)2αβ

On putting values of α+β and αβ from above we get:-

Sum of zeroes = α1 + β1 = 423

= 163

Now, Product of zeroes = (1+αβ)(1+βα)

=1+βα+αβ+αββα

=2αβ+β2+α2αβ

= (α+β)2αβ

On putting values of α+β and αβ we get:

Product of zeroes = α1 × β1 = 423

= 163

Thus the required quadratic polynomial equation will be:-

x2 – (α1 + β1)x + α1β1 = 0

x2 – (163)x + 163 = 0

3x2 – 16x +16=0.

 

Q4. If α and β are zeroes of a quadratic polynomial p(x) = rx2+4x+4, Find the values of “r” if: – α2 + β2 = 24.

Sol.

From the given polynomial p(x),

α + β = 4r, and αβ = 4r………. (1)

Since,(α + β)2 = α2 + β2 + 2αβ

Given,   α2 + β2 = 24 and from equation (1).

Therefore,  (4r)2=24+2×4r

16r2=24+2×4r

16 = 24 r2 + 8r

3r2 + r – 2 = 0

3r2 + 3r – 2r -2 = 0

3r(r+1) – 2(r+1) = 0

(3r-2) (r+1) = 0

∴  r = 23   or   r = -1

 

                                 DIVISION ALOGORITHM

If suppose p(x) and g(x) are any 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) …….(1)

This is known as the division algorithm for polynomials.

And since, Dividend = Divisor × Quotient + Remainder

On comparing it with equation (1) we can conclude that:

Dividend = p(x).

Divisor = g(x).

Quotient = q(x).

Remainder = r(x).

 

                              EXERCISE 2.3

Q.1 If a polynomial x3 -3x2+x+2 is divided by a polynomial g(x), the quotient and remainder obtained are (x-2) and (-2x+4), respectively. Find the equation of g(x).

Sol.

Since, Dividend = Divisor × Quotient + Remainder

Therefore, 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+3x-2

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

24

Therefore, g(x) = (x2 – x +1)

 

Q.2 Find the quotient and remainder by dividing the polynomial f(x) by the polynomial g(x).

(i) f(x) = x+ 2x– 9x + 5,   g(x) = x2+5

5

Therefore, Quotient is (x+2) and Remainder is (-14x − 5)

(ii) f(x) = x5+2x4-9x3+5x2-2x+1, g(x) = x3 + x2 – x+1

6

Therefore, Quotient is (x2 + x − 9) and Remainder is (14x2 – 12x +10).

 

(iii) f(x) = 2x+ 7x+ 5x+ 8x + 5, g(x) = 11− 2x3 + x2

7

Therefore, Quotient is  – (x + 4) and Remainder is (9x2 + 19x +49).

 

Q3. Find all the zeroes of the polynomial equation 2x4-3x3-3x2+6x-2, if two of its zeroes are 2and2.

Sol.

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

Let, f(x) = 2x– 3x– 3x+ 6x – 2

2and2 are zeroes of f(x).

(x2)(x+2)= (x2− 2) = g(x), is a factor of given polynomial f(x).

If we divide f(x) by g(x), the quotient will also be a factor of f(x) with remainder =0.

8

So, 2x4 − 3x3 − 3x2 + 6x – 2 = (x2 – 2) (2x2 – 3x +1).

Now, on further factorizing (2x2 – 3x +1) we get,

2x2 – 3x +1 = 2x2 – 2x − x +1 = 0

2x (x − 1) – 1(x−1) = 0

(2x−1) (x−1) = 0

So, its zeroes are given by:  x= 12 and x = 1

Therefore, all four zeroes of the given polynomial equation are:

2,2,12and1.

 

Q4. Find all zeroes of a polynomial equation x4-6x3-26x2+138x-35, if two of its zeroes are 2+3 and 23.

Sol.

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

Let, f(x) = x– 6x– 26x2+ 138x – 35

Since, (2+3)and(23) are zeroes of given polynomial f(x).

[x(2+3)][x(23)]=0

(x23)(x2+3)

On multiplying the above equation we get,

x2 −4x + 1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

9

So, x4 − 6x− 26x2 + 138x – 35 = (x2 – 4x + 1) (x2 – 2x −35).

Now, on further factorizing (x2 – 2x −35) we get,

x2 – (7−5)x − 35 = x2 – 7x + 5x +35 = 0

x(x − 7) + 5 (x−7) = 0

                                         (x+5) (x−7) = 0

So, its zeroes are given by:              x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are:

2+3,23, −5 and 7.

 

Q5. Find all zeroes of a polynomial equation 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are 53 and 53.

Sol.

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

Let, f(x)= 3x4+6x3-2x2-10x-5

53 and 53 are zeroes of polynomial f(x).

(x53)(x+53)

 (x253)=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 factor of f(x) and remainder will be 0.

10

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:

53, 53, −1 and −1.

Class 10 is an important phase of a student's life. The marks score in class 10 will help you to take admission in your preferred stream in class 11. The marks not only help in getting the preferred stream in class 11 but it also help in getting qualified for job openings. The Central Board of Secondary Education or CBSE is responsible for conducting the board examination of class 10th as well as for class 12th examination. The schools affiliated with CBSE are always advised to follow the NCERT curriculum as well as the NCERT textbooks, solving the NCERT questions provided at the end of each chapter will help the students to understand the chapter in a better way. Sometimes the NCERT questions are are directly asked in the board examinations, so solving the questions will help you to stay ahead from other students.

Students are slos advised to solve last five year questions of CBSE class 10th board examination as well as the sample papers. Solving one sample papers and one previous year questions everyday will help the students to know about the questions.

Keep visiting BYJU’S for latest CBSE notes and NCERT solutions for classes 6th to 12th.

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