Careers360 Logo
School
Search Colleges, Exams, Schools & more
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Edited By Apoorva Singh | Updated on Sep 04, 2023 09:03 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Class 10 Maths Chapter 2 Solutions Polynomials will discuss about the polynomials with any degree and its solutions using a graphical representation. These solutions are prepared by expert teachers keeping in mind the latest CBSE syllabus 2023. Also, Class 10 maths chapter 2 solutions polynomials gives a detailed explanation of questions given in the NCERT Book of Class 10 Maths.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
  2. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials PDF Free Download
  3. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials - Important Formulae
  4. Class 10 Maths Chapter 2 Solutions Polynomials (Intext Questions and Exercise)
  5. Polynomials Class 10 - Topics
  6. NCERT Books and NCERT Syllabus here
  7. NCERT Solutions for Class 10 Maths - Chapters Wise
  8. Features of NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
  9. NCERT Solutions of Class 10 - Subject Wise
  10. How to use NCERT solutions for class 10 maths chapter 2 polynomials?
  11. NCERT Exemplar solutions - Subject wise
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

You must refer to these Polynomial 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.

Apply to Aakash iACST Scholarship Test 2024

Applications for Admissions are open.
Apply Now

Also Read,

Tallentex 2025 - ALLEN's Talent Encouragement Exam

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

Apply
Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

Apply

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials PDF Free Download

Download PDF

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials - Important Formulae

The general Polynomial Expression is represented as follows:

  • F(x) = anxn + an-1xn-1 + an-2xn-2 + ... + rx + s.

For a natural number n, an-bn = (a - b)(an-1 + an-2b + ... + bn-2a + bn-1)

When n is an even number (expressed as n = 2a),

xn + yn = (x + y)(xn-1 - xn-2y + ... + yn-2x - yn-1)

For an odd number n,

xn + yn = (x + y)(xn-1 - xn-2y + ... - yn-2x + yn-1)

Different types of polynomials encompass various essential concepts and properties, each delineated in detail in the provided table.

1693828935548

Algebraic polynomial identity

  1. (a + b)2 = a2 + b2 + 2ab

  2. (a - b)2 = a2 + b2 - 2ab

  3. (a + b)(a - b) = a2 - b2

  4. (x + a)(x + b) = x2 + (a + b)x + ab

  5. (x + a)(x - b) = x2 + (a - b)x - ab

  6. (x - a)(x + b) = x2 + (b - a)x - ab

  7. (x - a)(x - b) = x2 - (a + b)x + ab

  8. (a + b)3 = a3 + b3 + 3ab(a + b)

  9. (a - b)3 = a3 - b3 - 3ab(a - b)

  10. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz

  11. (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz

  12. (x -y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz

  13. (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz

  14. x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

  15. x2 + y2 = 12[(x + y)2 + (x - y)2]

  16. (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

  17. x3 + y3 = (x + y)(x2 - xy + y2)

  18. x3 - y3 = (x - y)(x2 + xy + y2)

  19. x2 + y2 + z2 - xy - yz - zx = 1/2[(x - y)2 + (y - z)2 + (z - x)2]

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Apply

The division algorithm for polynomials states that when considering any two polynomials p(x) and g(x). Where g(x) is not equal to zero, Then we can find polynomials q(x) and r(x) such that:

p(x) = q(x) g(x) + r(x)

Here, r(x) = 0 or deg r(x) ≥ deg g(x)

Free download NCERT Solutions for Class 10 Maths Chapter 2 Polynomials PDF for CBSE Exam.

Class 10 Maths Chapter 2 Solutions Polynomials (Intext Questions and Exercise)


NCERT Solutions for Chapter 2 Maths Class 10 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.


Class 10 Maths Chapter 2 Solutions 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 Chapter 2 Maths Class 10 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

NCERT Books and NCERT Syllabus here

NCERT Solutions for Class 10 Maths - Chapters Wise

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.

Also See,

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.

NCERT Exemplar solutions - Subject wise

Frequently Asked Question (FAQs)

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

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.

2. List the topics that the NCERT Class 10 Chapter 2 Maths

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

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

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.

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

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.

5. What does chapter 2 Class 10 NCERT define as a polynomial?

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.

Also Read

Articles

Upcoming School Exams

National Means Cum-Merit Scholarship

Application Date:15 July,2024 - 14 August,2024

National Institute of Open Schooling 10th examination

Late Fee Application Date:22 July,2024 - 31 July,2024

National Institute of Open Schooling 12th Examination

Late Fee Application Date:22 July,2024 - 31 July,2024

Chhattisgarh Board of Secondary Education 12th Examination

Exam Date:23 July,2024 - 12 August,2024

Chhattisgarh Board of Secondary Education 10th Examination

Exam Date:24 July,2024 - 08 August,2024

View All School Exams

Certifications By Top Providers

Edx
 1111 courses
Coursera
 792 courses
Udemy
 327 courses
Futurelearn
 147 courses
Harvard University, Cambridge
 98 courses
IBM
 85 courses

Explore Top Universities Across Globe

University of Essex, Colchester
  Wivenhoe Park Colchester CO4 3SQ
University College London, London
  Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
  Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
  Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
  University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
  Bailrigg, Lancaster LA1 4YW

Related E-books & Sample Papers

CBSE 10th Hindi B syllabus 2024-25

144+ Downloads

CBSE 10th Hindi A syllabus 2024-25

71+ Downloads

CBSE 12th Hindi A syllabus 2024-25

102+ Downloads

Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer
Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Read Complete Answer
sadafreen356 03 July,2024
can we apply for diploma if we have passed 10th cbse lo

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.

You can pursue diplomas at various institutions like:


Read Complete Answer
Nitin Kumar 14 June,2024
View All

Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

Colleges After 12th

Applications for Admissions are open.

Tallentex 2025 - ALLEN's Talent Encouragement Exam
Tallentex 2025 - ALLEN's Talent Encouragement Exam
Apply

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

Aakash iACST Scholarship Test 2024
Aakash iACST Scholarship Test 2024
Apply

Get up to 90% scholarship on NEET, JEE & Foundation courses

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

PTE Exam 2024 Registrations
PTE Exam 2024 Registrations
Apply

Register now for PTE & Save 5% on English Proficiency Tests with ApplyShop Gift Cards

View All Application Forms
News and Notifications
July 03, 2024 - 12:56 PM IST :
June 27, 2024 - 01:32 PM IST :
CBSE: No open book exams for Classes 10, 12 this year; pilot only in a few schools
June 12, 2024 - 10:52 AM IST
CBSE Class 10 result 2024 revaluation application begins at cbse.gov.in; apply by May 24
May 21, 2024 - 02:20 PM IST
CBSE Class 10 Result 2024 (Out) LIVE: CBSE 10th results declared at cbseresults.nic.in; direct link
May 13, 2024 - 10:35 PM IST
CBSE 10th, 12th Results 2024: How to calculate percentage
May 13, 2024 - 03:22 PM IST
CBSE 10th result 2024 improves in government schools; less students placed in compartment
May 13, 2024 - 03:16 PM IST
CBSE 10th Result 2024: Highest number of students appear this year; 20.95 lakh pass exam
May 13, 2024 - 02:18 PM IST
CBSE Class 10th Result 2024 Out: Trivandrum top performing district again; girls outperform boys
May 13, 2024 - 01:41 PM IST

Stay up-to date with CBSE Class 10th News

Back to top