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    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.

    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.

    Also Read,

    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]

    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


    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 (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

    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.

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    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

    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

     

    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

    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

    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 .

    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

    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.

    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.

    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

    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

    Bio Medical Engineer

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    Also Read: Career as Nurse

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    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

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    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

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    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

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    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

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    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

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    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

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    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

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    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

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    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

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