NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Updated on Jan 15, 2021 - 9:37 a.m. IST by Apoorva Singh #CBSE Class 10th

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials - In this chapter, you will learn about the polynomials with any degree and its solutions using a graphical representation. The NCERT Solutions for Class 10 Maths Chapter 2 Polynomials gives a detailed explanation of questions given in the NCERT Book of Class 10 Maths. You must refer to these NCERT solutions to check the correct answers and proper explanation of each question. These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are prepared by experts and are easy understand. You can check the NCERT solutions for Class 10 for other subjects here.

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NCERT Solutions for Class 10 Maths Chapter 2 - Topics

The degree of a polynomial
Number of zeroes in a quadratic polynomial
The sum & product of quadratic polynomial
Division algorithm
The number of zeroes in a cubic polynomial

NCERT solutions for class 10 maths chapter 2 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.

sdfsdfs

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

s10

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

s11

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

s12

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

NCERT solutions for class 10 maths chapter 2 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:

x ² - 2x - 8 = 0

x ² - 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:

6x ² - 3 - 7x = 0

6x ² - 7x - 3 = 0

6x ² - 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: 4u ² + 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: t ² - 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: 3x ² - x - 4 = 0

3x ² + 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 (1) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1/4 , -1

Answer:

$\\\alpha +\beta =\frac{1}{4}\\ \alpha \beta =-1$

The required quadratic polynomial is

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

Q2 (2) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $\sqrt 2 , 1/3$

Answer:

$\\\alpha +\beta =\sqrt{2}\\ \alpha \beta =\frac{1}{3}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\sqrt{2}x+\frac{1}{3}=0\\ 3x^{2}-3\sqrt{2}x+1=0$

The required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$

Q2 (3) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $0,\sqrt{5}$

Answer:

$\\\alpha +\beta =0\\ \alpha \beta =\sqrt{5}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-0x+\sqrt{5}=0\\ x^{2}+\sqrt{5}=0$

The required quadratic polynomial is x ² + $\sqrt{5}$ .

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

Answer:

$\\\alpha +\beta =1\\ \alpha \beta =1$

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

The required quadratic polynomial is x ² - x + 1

Q2 (5) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $-\frac{1}{4} , \frac{1}{4}$

Answer:

$\\\alpha +\beta =-\frac{1}{4}\\ \alpha \beta =\frac{1}{4}$

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

The required quadratic polynomial is 4x ² + x + 1

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 x ² - 4x + 1

NCERT solutions for class 10 maths chapter 2 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

s15

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

s16

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

s17

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:

s18

After dividing we got the remainder as zero. So $t^2 - 3$ is 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

s20

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

s22

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

mp1

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 dividing $x^3 - 3x^2 + x + 2$ 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

s23

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

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

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

NCERT solutions for class 10 maths chapter 2 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) = 2x ³ + x ² -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 1 ³ + 1 ² - 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 ax ³ + bx ² + 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) = x ³ - 4x ² + 5x - 2

p(2) = 2 ³ - 4 x 2 ² + 5 x 2 - 2

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

p(-2) = 0

p(1) = 1 ³ - 4 x 1 ² + 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 ax ³ + bx ² + 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 x ³ - 2x ² - 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 - b ² = -1

b ² - 2 = 0

$b=\pm \sqrt{2}$

Therefore a = 1 and $b=\pm \sqrt{2}$ .

Q4 If two zeroes of the polynomial $x^4 - 6x ^3 - 26 x^2 + 138 x - 35$ $x^4 - 6x ^3 - 26 x^2 + 138 x - 35$ are $2 \pm \sqrt 3$ , 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

s26

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

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

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

s28

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 for other chapters

Chapter No.	Chapter Name
Chapter 1	NCERT solutions for class 10 maths chapter 1 Real Numbers
Chapter 2	NCERT solutions for class 10 maths chapter 2 Polynomials
Chapter 3	NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables
Chapter 4	NCERT solutions for class 10 maths chapter 4 Quadratic Equations
Chapter 5	NCERT solutions for class 10 chapter 5 Arithmetic Progressions
Chapter 6	NCERT solutions for class 10 maths chapter 6 Triangles
Chapter 7	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	NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
Chapter 10	NCERT solutions class 10 maths chapter 10 Circles
Chapter 11	NCERT solutions for class 10 maths chapter 11 Constructions
Chapter 12	NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles
Chapter 13	NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes
Chapter 14	NCERT solutions for class 10 maths chapter 14 Statistics
Chapter 15	NCERT solutions for class 10 maths chapter 15 Probability

Features of NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

The NCERT Class 10 maths solutions chapter 2 are important for board exams as well as for competitive exams.
These NCERT solutions are prepared by experts. These solutions are very well explained and are to the point.
To make the solutions easy to understand, relevant diagrams or images are given with the solutions wherever required.
You can check these solutions online or can save the webpage to get these solutions offline. NCERT Solutions for Class 10 Maths Chapter 2 PDF download will also be made available soon.

NCERT solutions of class 10 subject wise

How to use NCERT solutions for class 10 maths chapter 2 polynomials?

You will find yourself much improved in terms of concepts, their applications, and problem-solving. 90% of the questions in the board examinations come from the NCERT syllabus.
Now when you have gone through the NCERT solutions for class 10 maths chapter 2 polynomials, you must have learnt to answer a question in the step by step manner.
Once you have studied the Class 10 Maths chapter 2 NCERT solutions, your next target should be previous year questions papers.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After going through these two things, you can jump to the next chapters.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Question: What is the weightage of the chapter polynomials for CBSE board exam?

Answer:

CBSE does not provide the marks distributions chapter-wise but it provides the total weightage of a unit. As per CBSE the total weightage of algebra (4 chapters) is 20 marks in the final board exam.

Question: What are the important topics of the class 10 maths chapter 2 polynomials?

Answer:

Polynomial and it's definition, degree of a polynomial, number of zeroes in a quadratic polynomial, sum & product of quadratic polynomial, division algorithm, number of zeroes in a cubic polynomial are important topics from the chapter polynomials.

Question: How many chapters are there in class 10 maths?

Answer:

There are 15 chapters in the class 10 maths NCERT.

Question: How are the NCERT solutions helpful in the board exam?

Answer:

NCERT solutions are provided in a detailed manner, so it will be very easy for you to understand the concept. You can also use them for quick revision before the board exam.

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Age criteria to appear in class 10 CBSE exam

Manisha 31st Jan, 2021

Hello Bishal

According to guidelines of CBSE, minimum age to appear for class x must be 14 years. There is no upper limit to appear for class x cbse board.

A candidate can appear for maximum three attempts.

Some candidates give private exams or sometimes students fail in standard ix then they privately appear for class x then their age must be more than 14 years. Sometimes students appear for x class after one year gap of passing class ix then also their age would be 15 or 16 as there is no upper limit age.

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My child is supposed to make another attempt at compartment exam of 10th cbse Missed last date for applying. since the boards are postponed to May 4th can I pay her fees and apply now.

Yash Dev Student Expert 15th Jan, 2021

Hello sir I'm sorry to inform you but now you're not eligible to fill the registration form as last date to fill the registration Form was 9th December 2020. Yes examination gets postponed to May but portal to fill the registration form is not open , in case if the registration form portal will open again you can fill the registration form and make the payment.

Feel free to comment if you've any doubt

Good luck

144 Views

i want to know the sample paper and marking scheme of 10 th science

Parmeshwar Suhag 3rd Jan, 2021

Hello there,

To get the sample papers for CBSE Class 10th, you should visit the official CBSE website. There you can get all the papers. Solving the previous year papers, will be very beneficial for you. So, solve as much papers as you can.

As per the latest CBSE Class 10 exam pattern 2020-21, the theory paper will be of 80 marks, while 20 marks will be for internal assessment. To know more about exam pattern, please visit the link given below.

https://www.google.com/amp/s/school.careers360.com/articles/cbse-10th-exam-pattern/amp

Check out the papers here:

https://www.google.com/amp/s/school.careers360.com/articles/cbse-class-10-sample-papers/amp

753 Views

will the 2021 board exams for grade 10 conducted online or offline

Shubham Dhane 4th Nov, 2020

Hello,

The 10th CBSE board exam will be conducted in offline mode only, These board exams will not be conducted in online mode. Central Board of Secondary Education conducts class 10th board examination for regular and private students. The exams of CBSE 10th 2021 will be conducted in February and March 2021. Candidates can fill CBSE 10th application form 2021 online on cbse.nic.in from October 24 to November 11, 2020.

144 Views

is unit exam marks taken into consideration for cbse internal marks grade 10.

Sameer Kumar Thakur 1st Oct, 2020

Hey,

Your concern completely depends upon school. School management plans the division of marks. For few schools they must consider even class test on the other hand some just focus on mid and end . So the division of marks vary from school to school. What the school do is continuous access the student and then give marks accorrdingly. In my opinion focus on each and every test. Apart from internals it work for you in finals as well.

All the best

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2 - Topics

NCERT solutions for class 10 maths chapter 2 Polynomials Excercise: 2.1

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

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

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

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

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

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

NCERT solutions for class 10 maths chapter 2 Polynomials Excercise: 2.2

Answer:

Answer:

Answer: 4u 2 + 8u = 0

Answer: t 2 - 15 = 0

Answer: 3x 2 - x - 4 = 0

Answer:

Answer:

Answer:

Answer:

Answer:

NCERT solutions for class 10 maths chapter 2 Polynomials Excercise: 2.3

Answer: The polynomial division is carried out as follows

Answer:

Answer: The polynomial is divided as follows

Answer:

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

Answer: The polynomial division is carried out as follows

Answer: Quotient = x-2

Answer: 4u ² + 8u = 0

Answer: t ² - 15 = 0

Answer: 3x ² - x - 4 = 0

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