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

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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.

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials PDF Free Download
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials - Important Formulae
Class 10 Maths Chapter 2 Solutions Polynomials (Intext Questions and Exercise)
Polynomials Class 10 - Topics
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
NCERT Solutions of Class 10 - Subject Wise
How to use NCERT solutions for class 10 maths chapter 2 polynomials?
NCERT Exemplar solutions - Subject wise

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.

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

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

The general Polynomial Expression is represented as follows:

F(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ... + rx + s.

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

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

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

For an odd number n,

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

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

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Algebraic polynomial identity

(a + b)² = a² + b² + 2ab
(a - b)² = a² + b² - 2ab
(a + b)(a - b) = a² - b²
(x + a)(x + b) = x² + (a + b)x + ab
(x + a)(x - b) = x² + (a - b)x - ab
(x - a)(x + b) = x² + (b - a)x - ab
(x - a)(x - b) = x² - (a + b)x + ab
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - b³ - 3ab(a - b)
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz
(x + y - z)² = x² + y² + z² + 2xy - 2yz - 2xz
(x -y + z)² = x² + y² + z² - 2xy - 2yz + 2xz
(x - y - z)² = x² + y² + z² - 2xy + 2yz - 2xz
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
x² + y² = 12[(x + y)² + (x - y)²]
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
x³ + y³ = (x + y)(x² - xy + y²)
x³ - y³ = (x - y)(x² + xy + y²)
x² + y² + z² - xy - yz - zx = 1/2[(x - y)² + (y - z)² + (z - x)²]

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

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

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These NCERT solutions are prepared by experts. These solutions are very well explained and are to the point.
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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 Questions (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

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer

sai raja 01 September,2024

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

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Central Board of Secondary Education Class 10th Examination

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Pearson | PTE

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

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

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

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.

Answer:

Answer:

Answer: 4u 2 + 8u = 0

Answer: t 2 - 15 = 0

Answer: 3x 2 - x - 4 = 0

Answer:

Answer:

Answer:

Answer:

Answer:

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

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

Answer:

Answer:

Answer:

Answer: Let the roots of the polynomial be

Answer:

Answer: Given the two zeroes are

Answer: The polynomial division is carried out as follows

Polynomials Class 10 - Topics

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

NCERT Solutions of Class 10 - Subject Wise

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

NCERT Exemplar solutions - Subject wise

Frequently Asked Questions (FAQs)

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Answer: 4u ² + 8u = 0

Answer: t ² - 15 = 0

Answer: 3x ² - x - 4 = 0

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