NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Edited By Apoorva Singh | Updated on Feb 8, 2023 - 12:34 p.m. IST | #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.

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

You must refer to these Polynomials Class 10 notes 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 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.

1635918537229

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

1635918562539

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

1635918572380

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

1635918582058

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

1635918596067

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

1635918606641

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

1635918804777

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

1635919021607

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

1635919038492

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:

1635919067387

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

1635919079452

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

1635919099878

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

1635919117816

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

1635919149497

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

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

1635919283320

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

1635919302615

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

Polynomials Class 10 - 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

Also See,

NCERT Solutions for Class 10 Maths for other chapters

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Features of NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

The chapter 2 Maths Class 10 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 Class 10 Maths chapter 2 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 Maths chapter 2 Polynomials Class 10 , 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.

Subject wise NCERT Exemplar solutions

Also Check NCERT Books and NCERT Syllabus here

Solutions

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

Question: What does chapter 2 Class 10 NCERT define as a polynomial?

Answer:

According to the NCERT chapter 2 Class 10 Maths Polynomials, a polynomial is a mathematical expression that consists of variables (such as x, y, etc.) and coefficients (numbers), which are combined using only the operations of addition, subtraction, and multiplication.

Question: How many Exercises are there in Chapter 2 of Polynomials in the NCERT Solutions for Class 10 Mathematics?

Answer:

There are a total of 10 exercises in Chapter 2 (Polynomials) of the NCERT Solutions for Class 10 Mathematics. These exercises consist of a variety of questions, including multiple choice questions, fill in the blanks, true or false, and short answer questions. The number of questions in each exercise may vary.

Question: Why Should You Use the NCERT Solutions for Chapter 2 of Math in Class 10?

Answer:

There are several reasons why you should use the ncert solutions for class 10 maths chapter 2 polynomials:

The NCERT Solutions are accurate and reliable
The NCERT Solutions cover all the concepts and exercises given in the textbook
The NCERT Solutions can help you develop problem-solving skills
The NCERT Solutions can be used as a reference guide

Overall, using the polynomials class 10 questions with answers can be a useful resource for students to improve their understanding of the concepts and perform well in exams.

Question: List the topics that the NCERT Class 10 Chapter 2 Maths

Answer:

The topics covered in NCERT maths class 10 chapter 2 (Polynomials) are:

Definition and types of polynomials
Zeroes of a polynomial
Division algorithm for polynomials
Fundamental Theorem of Algebra
Relationship between zeroes and coefficients of a polynomial
Nature of roots of a polynomial

Question: Where can I find the correct answer to Maths NCERT Chapter 2 polynomials class 10 Solutions?

Answer:

At the careers360 official website, you can find polynomials class 10 ncert solutions. These solutions for ch2 maths class 10 are designed and developed by reliable experts keeping in mind the new pattern provided by the state and central boards. Also, you can download it as a polynomials class 10 pdf.

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Questions related to CBSE Class 10th

Showing 71 out of 71 Questions

404 Views

my sons dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

Shubhangi 30th Jul, 2022

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

1467 Views

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

Aditi Singh 22nd Jul, 2022

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

134 Views

i got less marks in maths can it get replaced by additional hindi in cbse

Aditi Singh 22nd Jul, 2022

Replacement of marks of additional subjects and your main course subject is not like they will swap the marks you got in Hindi and Mathematics on your marksheet. It works like if you calculate the total percentage you got in your class, you can take the marks of your additional subject to consider in place of the subject you have got the least marks in. CBSE schools consider this method only to release the merit list of their students.

If you're in 10th and aiming to get admission in 11th in any school, it depends on the schools if they consider your main five subject marks or the best five subject marks for the admission.

If you're in 12th and are trying for different colleges to get in, it depends on your course which you wish to study further and also some criterias and eligibility of the colleges. In most cases for general courses, colleges take the best three or best four subjects marks consideration for the admission process.

So, good luck.

93 Views

when will cbse term 2 result will declare please tale

Shubham Dhane 21st Jul, 2022

Hello,

Central Board of Secondary Education will declare term 2 CBSE exam result 2022 for Class 10 in the third week of July. Please keep an eye on the official website or the link below for the latest information. Students need to enter their CBSE board roll number and other details to check term 2 cbseresults.nic.in 2022 Class 10 results. The board had announced the CBSE Class 10 results 2022 for term 1 on March 11, 2022.

https://school.careers360.com/boards/cbse/cbse-result

51 Views

cbse 10 claas bord result check now

Jyoti Eyasmin 20th Jul, 2022

Hello Aspirant,

Class 10 cbse results are not out yet. It is expected to be declared on 23rd of July. Keep checking the official website of cbse.

Good luck.

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NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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: 4u ² + 8u = 0

Answer: t ² - 15 = 0

Answer: 3x ² - x - 4 = 0

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