NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

 

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials- In this particular chapter you will study the geometrical interpretation of zeroes of polynomial and how to carry out divisions of the polynomials. In mathematics, there are two kinds of numbers- one is the real number and another one is the imaginary numbers. NCERT solutions for class 10 maths chapter 1 real number has very detailed solutions to each and every question. This chapter will be dedicated to real numbers. Real numbers are those numbers that can be represented in the number line. It is one of the most elegant topics in NCERT class 10 maths which can be connected to real life. If anyone wants to go deeper into the number system then the real number is the first step of a mastering number system. Real numbers include the in-depth classification of numbers and applications and properties related to various kinds of numbers. The number system is also very important when you will be appearing in the competitive examinations. CBSE NCERT solutions for class 10 maths chapter 1 real numbers will give you a great approach to the application of a concept in the particular question.

Types of questions asked from class 10 maths chapter 2 Polynomials-

CBSE Class 10 maths board exam will have the following types of questions from the chapter polynomials:

  • Questions related to the degree of a polynomial

  • Questions based on the number of zeroes in a quadratic polynomial

  • Questions related to the sum & product of quadratic polynomial

  • Questions based on division algorithm

  • Questions on the number of zeroes in a cubic polynomial

Importance of Chapter 1 Real Numbers in Class 10 Maths

What is the use of Polynomials?. Polynomials are used in various fields of Science, for analysis purposes and in the field of economics for cost analysis, etc.  In our daily life, we use polynomials for calculations frequently. For example, The length of a rectangle is 7 cm more than the breadth and the perimeter of the rectangle is 21 cm. Find the breadth and length? This can be solved by formulating a polynomial as x+(x+7)=21. By solving this we will get length as 14cm and breadth as 7cm. Every year in the board examinations. There are always 1-2 questions carrying 6 to 7 marks that come from this chapter.

Polynomials Excercise: 2.1

Q1 (1) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the numbers of zeroes of p(x), in each case.

               

Answer:

The number of zeroes of p(x) is zero as the curve does not intersect the x-axis.

Q1 (2) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case 

               

Answer:

The number of zeroes of p(x) is one as the graph intersects the x-axis only once.

Q1 (3) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

            

 

Answer:

The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

Q1 (4) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

             

Answer:

The number of zeroes of p(x) is two as the graph intersects the x-axis twice.

Q1 (5) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

             

Answer:

The number of zeroes of p(x) is four as the graph intersects the x-axis four times.

Q1 (6) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

            

Answer:

The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

Polynomials Excercise: 2.2

Q1 (1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.  x ^2 - 2x -8

Answer:

x2 - 2x - 8 = 0

x2 - 4x + 2x - 8 = 0

x(x-4) +2(x-4) = 0

(x+2)(x-4) = 0

The zeroes of the given quadratic polynomial are -2 and 4

\\\alpha =-2\\ \beta =4

VERIFICATION

Sum of roots:

\\\alpha+\beta =-2+4=2

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-2}{1}\\ =2\\=\alpha +\beta

Verified

Product of roots:

\\\alpha \beta =-2\times 4=-8

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-8}{1}\\ =-8\\=\alpha \beta

Verified

Q1 (ii) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4s^2 - 4s + 1

Answer:

\\4s^2 - 4s + 1 = 0 \\4s^2 - 2s - 2s + 1 = 0 \\2s(2s-1) - 1(2s-1) = 0 \\(2s-1)(2s-1) = 0

The zeroes of the given quadratic polynomial are 1/2 and 1/2

\\\alpha =\frac{1}{2}\\ \beta =\frac{1}{2}

VERIFICATION

Sum of roots:

\\\alpha +\beta =\frac{1}{2}+\frac{1}{2}=1

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-4}{4}\\ =1\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{1}{4}\\ =\alpha \beta

Verified

Q1 (3)  Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 6x ^2 - 3 -7x

Answer:

6x2 - 3 - 7x = 0

6x2 - 7x - 3 = 0

6x2 - 9x + 2x - 3 = 0

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

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

The zeroes of the given quadratic polynomial are -1/3 and 3/2

\\\alpha =-\frac{1}{3}\\ \beta =\frac{3}{2}

Sum of roots:

\\\alpha+\beta =-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-7}{6}\\ =\frac{7}{6}\\=\alpha +\beta

Verified

Product of roots:

\\\alpha \beta =-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-3}{6}\\ =-\frac{1}{2}\\=\alpha \beta

Verified

Q1 (4) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.4 u^2 + 8u

Answer:

4u2 + 8u = 0

4u(u + 2) = 0

The zeroes of the given quadratic polynomial are 0 and -2

\\\alpha =0\\ \beta =-2

VERIFICATION

Sum of roots:

\\\alpha +\beta =0+(-2)=-2

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{8}{4}\\ =-2\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =0\times-2=0

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{0}{4}\\=0\\ =\alpha \beta

Verified

Q1 (5) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. t^2 - 15

Answer:

t2 - 15 = 0

(t-\sqrt{15})(t+\sqrt{15})=0

The zeroes of the given quadratic polynomial are -\sqrt{15} and \sqrt{15}

\\\alpha =-\sqrt{15}\\ \beta =\sqrt{15}

VERIFICATION

Sum of roots:

\\\alpha +\beta =-\sqrt{15}+\sqrt{15}=0

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{0}{1}\\ =0\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =-\sqrt{15}\times \sqrt{15}=-15

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-15}{1}\\=-15\\ =\alpha \beta

Verified

Q1 (6) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 3 x^2 - x - 4

Answer:

3x2 - x - 4 = 0

3x2 + 3x - 4x - 4 = 0

3x(x + 1) - 4(x + 1) = 0

(3x - 4)(x + 1) = 0

The zeroes of the given quadratic polynomial are 4/3 and -1

\\\alpha =\frac{4}{3}\\ \beta =-1

VERIFICATION

Sum of roots:

\\\alpha +\beta =\frac{4}{3}+(-1)=\frac{1}{3}

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-1}{3}\\ =\frac{1}{3}\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =\frac{4}{3}\times -1=-\frac{4}{3}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-4}{3}\\ =\alpha \beta

Verified

Q2 (6)  Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 4,1

Answer:

\\\alpha +\beta =4\\ \alpha \beta =1

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-4x+1=0\\

The required quadratic polynomial is x2 - 4x + 1

Polynomials Excercise: 2.3

Q1 (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 + 5x - 3, g(x) = x^2 - 2

Answer:

The polynomial division is carried out as follows

The quotient is x-3 and the remainder is 7x-9

Q1 (2) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : p(x) = x^4 - 3x^2 + 4x + 5, g(x) = x^2 + 1 - x

Answer:

The division is carried out as follows

The quotient is x^2+x-3

and the remainder is 8

Q1 (3) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
p(x) = x^4 - 5x + 6, g(x) = 2 - x^2

Answer:

 The polynomial is divided as follows

The quotient is -x^2-2 and the remainder is -5x+10

Q2 (1) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

              t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12

Answer:

After dividing we got the remainder as zero. So t^2 - 3is a factor of 2t^4 + 3t^3 + -2t^2 -9t -12

Q2 (2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

            x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2

Answer:

To check whether the first polynomial is a factor of the second polynomial we have to get the remainder as zero after the division

After division, the remainder is zero thus x^2+3x+1 is a factor of 3x^4 + 5x^3 -7x^2+2x+2

Q2 (3) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1

Answer:

The polynomial division is carried out as follows

The remainder is not zero, there for the first polynomial is not a factor of the second polynomial

Q3 Obtain all other zeroes of 3x^4 + 6x^3 - 2x^2 - 10x - 5, if two of its zeroes are \sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}

Answer:

Two of the zeroes of the given polynomial are \sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}.

Therefore two of the factors of the given polynomial are x-\sqrt{\frac{5}{3}} and x+\sqrt{\frac{5}{3}}

(x+\sqrt{\frac{5}{3}})\times (x-\sqrt{\frac{5}{3}})=x^{2}-\frac{5}{3}

x^{2}-\frac{5}{3}   is a factor of the given polynomial.

To find the other factors we divide the given polynomial with 3\times (x^{2}-\frac{5}{3})=3x^{2}-5

 

The quotient we have obtained after performing the division is x^{2}+2x+1

\\x^{2}+2x+1\\ =x^{2}+x+x+1\\ =x(x+1)+(x+1)\\ =(x+1)^{2}

(x+1)2 = 0

x = -1

The other two zeroes of the given polynomial are -1.

Q4 On dividingx^3 - 3x^2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Answer:

Quotient = x-2

remainder =-2x+4

\\dividend=divisor\timesquotient+remainder\\x^3-3x^2+x+2=g(x)(x-2)+(-2x+4)\\x^3-3x^2+x+2-(-2x+4)=g(x)(x-2)\\x^3-3x^2+3x-2=g(x)(x-2)\\

g(x)=\frac{x^3-3x^2+3x-2}{x-2}

Carrying out the polynomial division as follows

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

Q5 (1) 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)

Answer:

deg p(x) will be equal to the degree of q(x) if the divisor is a constant. For example

\\p(x)=2x^2-2x+8\\q(x)=x^2-x+4\\g(x)=2\\r(x)=0

Q5 (2) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg q(x) = deg r(x)

Answer:

Example for a polynomial  with deg q(x) = deg r(x) is given below

\\p(x)=x^3+x^2+x+1\g(x)=x^2-1\\q(x)=x+1\\r(x)=2x+2

Q 5 (3) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and  deg r(x) = 0

Answer:

example for the polynomial which satisfies the division algorithm with r(x)=0 is given below

\\p(x)=x^3+3x^2+3x+5\\q(x)=x^2+2x+1\\g(x)=x+1\\r(x)=4

Polynomials Excercise: 2.4

Q1 (1) Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:

2x^3 + x^2 - 5x +2 ; \frac{1}{2} , 1 , -2

Answer:

p(x) = 2x3 + x2 -5x + 2

\\p(\frac{1}{2})=2\times \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{2}-5\times \frac{1}{2}+2\\ p(\frac{1}{2})=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\\ p(\frac{1}{2})=0

p(1) = 2 x 13 + 12 - 5 x 1 + 2

p(1) =2 + 1 - 5 + 2

p(1) = 0

p(-2) = 2 x (-2)3 + (-2)2 - 5 x (-2) +2

p(-2) = -16 + 4 + 10 + 2

p(-2) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax3 + bx2 + cx + d, we have

a = 2, b = 1, c = -5, d = 2

The roots are \alpha ,\beta \ and\ \gamma

\\\alpha=\frac{1}{2}\\ \beta =1\\ \gamma =-2

\\\alpha+\beta +\gamma \\ =\frac{1}{2}+1+(-2)\\ =-\frac{1}{2}\\ =-\frac{b}{a}

Verified

\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}\\ =\frac{-5}{2}\\ =\frac{c}{a}

Verified

\\\alpha\beta\gamma \\=\frac{1}{2}\times 1\times -2\\ =-1 \\=-\frac{2}{2}\\ =-\frac{d}{a}

Verified

Q1 (2) Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

x ^ 3- 4x ^ 2 + 5x - 2 ; 2,1,1

Answer:

p(x) = x3 - 4x2 + 5x - 2

p(2) = 23 - 4 x 22 + 5 x 2 - 2

p(2) = 8 - 16 + 10 - 2

p(-2) = 0

p(1) = 13 - 4 x 12 + 5 x 1 - 2

p(1) = 1 - 4 + 5 - 2

p(1) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax3 + bx2 + cx + d, we have

a = 1, b = -4, c = 5, d = -2

The roots are \alpha ,\beta \ and\ \gamma

\\\alpha=2\\ \beta =1\\ \gamma =1

\\\alpha+\beta +\gamma \\ =2+1+1\\ =4\\ =-\frac{-4}{1}\\=-\frac{b}{a}

Verified

\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =2\times 1+1\times 1+1\times 2\\ =5\\ =\frac{5}{1}\\ =\frac{c}{a}

Verified

\\\alpha\beta\gamma \\=2\times 1\times 1\\=2 \\=-\frac{-2}{1}\\=-\frac{d}{a}

Verified

Q2 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Answer:

Let the roots of the polynomial be \alpha ,\beta \ and\ \gamma

\\\alpha +\beta +\gamma =2\\ \alpha \beta +\beta \gamma +\gamma \alpha =-7\\ \alpha \beta \gamma =-14

\\x^{3}-(\alpha +\beta +\gamma )x^{2}+(\alpha \beta +\beta \gamma +\gamma \alpha)x-\alpha \beta \gamma =0\\ x^{3}-2x^{2}-7x+14=0

Hence the required cubic polynomial is x3 - 2x2 - 7x + 14 = 0

Q3 If the zeroes of the polynomial x^3 - 3 x^2+ x +1  are a – b, a, a + b, find a and b.

Answer:

x^3 - 3 x^2+ x +1

The roots of the above polynomial are a, a - b and a + b

Sum of the roots of the given plynomial = 3

a + (a - b) + (a + b) = 3

3a = 3

a = 1

The roots are therefore 1, 1 - b and 1 + b

Product of the roots of the given polynomial = -1

1 x (1 - b) x (1 + b) = - 1

1 - b2 = -1

b- 2 = 0

b=\pm \sqrt{2}

Therefore a = 1 and b=\pm \sqrt{2}.

Q4 If two zeroes of the polynomial x^4 - 6x ^3 - 26 x^2 + 138 x - 35x^4 - 6x ^3 - 26 x^2 + 138 x - 35are &nbnbsp;2 \pm \sqrt 3 , find other zeroes.

Answer:

Given the two zeroes are

 2+\sqrt{3}\ and\ 2-\sqrt{3}

therefore the factors are

[x-(2+\sqrt{3})]\ and[x-(\ 2-\sqrt{3})]

We have to find the remaining two factors. To find the remaining two factors we have to divide the polynomial with the product of the above factors

[x-(2+\sqrt{3})]\[x-(\ 2-\sqrt{3})]=x^2-(2+\sqrt{3})x-(\ 2-\sqrt{3})x+(\ 2+\sqrt{3})(\ 2-\sqrt{3})\\=x^2-2x-\sqrt{3}x-2x+\sqrt{3}x+1\\=x^2-4x+1

Now carrying out the polynomial division

Now we getx^2-2x -35 \ is \ also \ a\ factor

\\x^2-2x -35 =x^2-7x+5x-35\\=x(x-7)+5(x-7)\\=(x-7)(x+5)

So the zeroes are 2+\sqrt{3}\ ,\ 2-\sqrt{3},7\ and\ -5

Q5 If the polynomial x ^4 - 6x ^3 + 16 x^2 - 35 x + 10  is divided by another polynomial x^2 - 2x + k, the remainder comes out to be x + a, find k and a.

Answer:

The polynomial division is carried out as follows

 Given the remainder =x+a

The obtained remainder after division is (2k-9)x+10-k(8-k)

now equating the coefficient of x

2k-9=1

which gives the value of k=5

now equating the constants

a=10-k(8-k)=10-5(8-5)=-5

Therefore k=5 and a=-5

NCERT solutions for class 10 maths - chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 chapter 5 Arithmetic Progressions

Chapter 6

Solutions of NCERT class 10 maths chapter 6 Triangles

Chapter 7

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

CBSE NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions  for class 10 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 10 maths chapter 15 Probability

Solutions of NCERT class 10 - subject wise

NCERT solutions for class 10 maths

Solutions of NCERT for class 10 science

Once you go through all the solutions from CBSE NCERT class 10 maths chapter 2 polynomials, you will find yourself much improved in terms of concepts, concept application, and problem-solving. 90% of the questions in the board examinations come from the NCERT textbook. If you are well prepared these 90% questions then the rest 10% can be fetched by doing some extension of what you have learned in terms of concepts.

Keep working hard & happy learning!

 

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