NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Updated on Aug 4, 2020 - 9:52 p.m. IST by Ravindra Pindel #CBSE Class 10th

NCERT solutions for class 10 maths chapter 2 Polynomials - In the previous class, you have already studied the polynomial with one degree. In class 10 maths chapter 2 you will study the polynomials with any degree and its solutions using graphical representation. NCERT solutions for class 10 maths chapter 2 Polynomials have detailed explanations to each and every question. A polynomial is constructed using variables and numbers using addition, subtraction, multiplication, and division. When two polynomials are divided then the resultant is called the rational expression of polynomials. NCERT solutions for class 10 maths chapter 2 Polynomials is an important tool to score high in the board examinations. Polynomial is an important part of algebra in which we get the knowledge about the various operation performed over polynomials. Polynomials are further classified according to their degrees. A degree in the polynomials is the highest power of variable present in the equation. Let's understand the degree by taking some examples-

In the equations, $x+1, x^{2}+2x+3$ the maximum power of the variables is 1 and 2 respectively. So the degree for the polynomials is 1 and 2 respectively. Polynomials with degrees 1, 2, 3 are known as linear, quadratic and cubic polynomials respectively. NCERT solutions for class 10 maths chapter 2 Polynomials is covering each question in a detailed manner. Apart from this particular chapter, you can also download the solutions class-wise and subject wise by clicking on the link where NCERT solutions are there. Here you will get NCERT solutions for class 10 also.

Types of questions asked from class 10 maths chapter 2 Polynomials

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.

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.

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

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 $t^2 - 3$ 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

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 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

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

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

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

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	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

NCERT solutions of class 10 subject wise

NCERT solutions for class 10 maths

NCERT solutions for class 10 science

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 NCERT solutions for class 10, 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.

Keep working hard & happy learning!

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 doesn't provides the marks distributions chapter-wise but it provides the total weightage of a unit( upto 4-5 chapters). As per CBSE the total weightage of algebra ( 4 chapters) is 20 marks in the final board exam.

Question: Where can I find the complete solutions of NCERT class 10 maths ?

Answer:

Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.

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: Which is the official website of NCERT ?

Answer:

http://ncert.nic.in/ is the official website of the NCERT where you can get NCERT textbooks from class 1 to 12 and syllabus from class 1 to 12 for all the subjects.

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

Answer:

There are 15 chapters in the class 10 maths NCERT. 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 are the chapters in the NCERT class 10 maths.

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

Answer:

NCERT solutions are very important, if have stuck into any problem while solving them. These solutions are provided in a detailed manners, 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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Questions related to CBSE Class 10th

Showing 21 out of 22 Questions

15 Views

I got 85% in CBSE class 10th , I am from open category. on the basis of merit list can I get admission in government polytechnic college, Aurangabad for computer science

Kamal Kishor Chaurasiya 30th Jul, 2020

Government Polytechnic College Aurangabad is a public college which was established in the year 1955. It has student capacity of 2042 students and has 87 faculty members.

This college offers 11 courses in 2 streams and 8 specializations. You have scored excellent in your board examination. Therefore you can surely get admission here.

For more information please check the link below: https://www.careers360.com/colleges/government-polytechnic-aurangabad/amp

19 Views

but I think I can rewrite the paper

Aishwary Jaiswal 22nd Jul, 2020

Hello Aspirant,

If you are talking about the CBSE class 10th exams and you have already appeared for the exams and are willing to re-appear . The reason mentioned by you is not clear and so I am supposing that this is because , you have got less/unexpected marks in your exams. To be honest, it's really difficult for you to rewrite the paper in this pandemic situation where every school and college is closed yet.

I would ask you to check out once our country deals effectively with the virus and the vaccine is out, then only something can be said with surety, until then it's just the assumptions.

Hope that helps

Stay safe

All the best.

12 Views

I not satisfied in only one subject is this wrong or correct to go for re-vaulution

Deblina Roy 16th Jul, 2020

Dear Student,

If you are totally confident about this subject then go for re-valuation. At first talk to your school or college authority about this and after being sure and knowing all the things then apply for this. May it will help you.

Thank You.

35 Views

CBSE class 10th compartment online forms Apply 2020

Shubham Dhane 16th Jul, 2020

Hello,

CBSE Class 10th compartment result is provided through the official website: cbseresults.nic.in. Students failing in a subject in annual exams of CBSE will be able to apply for compartment exam in the third week of August 2020. Last date to submit CBSE 10th compartment form is third week of August 2020. Please click the below link to know more.

https://school.careers360.com/articles/cbse-compartment-result-10th-class

106 Views

result 10th cbse ka chahiye or uska marksheet bhee

S. Saha Student Expert 15th Jul, 2020

CBSE has announced class 10 results 2020. Now, you can check the results on the official websites- cbseresults.nic.in or cbse.nic.in.

Go to the website.
Click on CBSE 10th result 2020.
Enter your credentials like roll number, school no., enter centre no., admit card number etc.
Click on submit button.
Your result will be displayed on the screen. Take a screenshot of your result.

You can also check your result through SMS.

Open SMS application.
Type SMS: cbse10<space>RollNumber<space>AdmitCardID
Send it to 7738299899.

Your CBSE result 2020 Class 10 will be sent to your mobile.
Students can also check their results through Umang and DigiLocker app. http://news.careers360.com/cbse-10th-result-2020-live-updates-check-cbse-class-10-results-cbseresultsnicin

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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

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

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

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