# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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

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

## Types of questions asked from class 10 maths chapter 2 Polynomials

• The degree of a polynomial

• Number of zeroes in a quadratic polynomial

• The sum & product of quadratic polynomial

• Division algorithm

• The number of zeroes in a cubic polynomial

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

The number of zeroes of p(x) is zero as the curve does not intersect the x-axis.

The number of zeroes of p(x) is one as the graph intersects the x-axis only once.

The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

The number of zeroes of p(x) is two as the graph intersects the x-axis twice.

The number of zeroes of p(x) is four as the graph intersects the x-axis four times.

The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

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

x2 - 2x - 8 = 0

x2 - 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

$\\4s^2 - 4s + 1 = 0 \\4s^2 - 2s - 2s + 1 = 0 \\2s(2s-1) - 1(2s-1) = 0 \\(2s-1)(2s-1) = 0$

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

6x2 - 3 - 7x = 0

6x2 - 7x - 3 = 0

6x2 - 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

4u2 + 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

t2 - 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

3x2 - x - 4 = 0

3x2 + 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

$\\\alpha +\beta =\frac{1}{4}\\ \alpha \beta =-1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\frac{1}{4}x-1=0\\ 4x^{2}-x-4=0$

$\\\alpha +\beta =\sqrt{2}\\ \alpha \beta =\frac{1}{3}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\sqrt{2}x+\frac{1}{3}=0\\ 3x^{2}-3\sqrt{2}x+1=0$

The required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$

$\\\alpha +\beta =0\\ \alpha \beta =\sqrt{5}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-0x+\sqrt{5}=0\\ x^{2}+\sqrt{5}=0$

The required quadratic polynomial is x2 + $\sqrt{5}$.

$\\\alpha +\beta =1\\ \alpha \beta =1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-1x+1=0\\ x^{2}-x+1=0$

The required quadratic polynomial is x2 - x + 1

$\\\alpha +\beta =-\frac{1}{4}\\ \alpha \beta =\frac{1}{4}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\left (-\frac{1}{4} \right )x+\frac{1}{4}=0\\ 4x^{2}+x+1=0$

The required quadratic polynomial is 4x2 + x + 1

$\\\alpha +\beta =4\\ \alpha \beta =1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-4x+1=0\\$

The required quadratic polynomial is x2 - 4x + 1

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

The polynomial division is carried out as follows

The quotient is x-3 and the remainder is 7x-9

The division is carried out as follows

The quotient is $x^2+x-3$

and the remainder is 8

The polynomial is divided as follows

The quotient is $-x^2-2$ and the remainder is $-5x+10$

$t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12$

After dividing we got the remainder as zero. So $t^2 - 3$ is a factor of $2t^4 + 3t^3 - 2t^2 - 9t - 12$

$x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$

To check whether the first polynomial is a factor of the second polynomial we have to get the remainder as zero after the division

After division, the remainder is zero thus$x^2+3x+1$ is a factor of $3x^4 + 5x^3 - 7x^2 + 2x + 2$

Q2 (3) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$

The polynomial division is carried out as follows

The remainder is not zero, there for the first polynomial is not a factor of the second polynomial

## Q3 Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeroes are $\sqrt {\frac{{ 5 }}{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$

Two of the zeroes of the given polynomial are $\sqrt {\frac{{ 5 }}{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$.

Therefore two of the factors of the given polynomial are $x-\sqrt{\frac{5}{3}}$ and $x+\sqrt{\frac{5}{3}}$

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

$x^{2}-\frac{5}{3}$   is a factor of the given polynomial.

To find the other factors we divide the given polynomial with $3\times (x^{2}-\frac{5}{3})=3x^{2}-5$

The quotient we have obtained after performing the division is $x^{2}+2x+1$

$\\x^{2}+2x+1\\ =x^{2}+x+x+1\\ =x(x+1)+(x+1)\\ =(x+1)^{2}$

(x+1)2 = 0

x = -1

The other two zeroes of the given polynomial are -1.

Quotient = x-2

remainder =-2x+4

$\\dividend=divisor\timesquotient+remainder\\x^3-3x^2+x+2=g(x)(x-2)+(-2x+4)\\x^3-3x^2+x+2-(-2x+4)=g(x)(x-2)\\x^3-3x^2+3x-2=g(x)(x-2)\\$

$g(x)=\frac{x^3-3x^2+3x-2}{x-2}$

Carrying out the polynomial division as follows

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

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

$\\p(x)=2x^2-2x+8\\q(x)=x^2-x+4\\g(x)=2\\r(x)=0$

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

$\\p(x)=x^3+x^2+x+1\g(x)=x^2-1\\q(x)=x+1\\r(x)=2x+2$

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:

$2x^3 + x^2 - 5x +2 ; \frac{1}{2} , 1 , -2$

p(x) = 2x3 + x2 -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 13 + 12 - 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 ax3 + bx2 + 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

$x ^ 3- 4x ^ 2 + 5x - 2 ; 2,1,1$

p(x) = x3 - 4x2 + 5x - 2

p(2) = 23 - 4 x 22 + 5 x 2 - 2

p(2) = 8 - 16 + 10 - 2

p(-2) = 0

p(1) = 13 - 4 x 12 + 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 ax3 + bx2 + 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

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 x3 - 2x2 - 7x + 14 = 0

$x^3 - 3 x^2+ x +1$

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 - b2 = -1

b- 2 = 0

$b=\pm \sqrt{2}$

Therefore a = 1 and $b=\pm \sqrt{2}$.

Given the two zeroes are

$2+\sqrt{3}\ and\ 2-\sqrt{3}$

therefore the factors are

$[x-(2+\sqrt{3})]\ and[x-(\ 2-\sqrt{3})]$

We have to find the remaining two factors. To find the remaining two factors we have to divide the polynomial with the product of the above factors

$[x-(2+\sqrt{3})]\[x-(\ 2-\sqrt{3})]=x^2-(2+\sqrt{3})x-(\ 2-\sqrt{3})x+(\ 2+\sqrt{3})(\ 2-\sqrt{3})\\=x^2-2x-\sqrt{3}x-2x+\sqrt{3}x+1\\=x^2-4x+1$

Now carrying out the polynomial division

Now we get$x^2-2x -35 \ is \ also \ a\ factor$

$\\x^2-2x -35 =x^2-7x+5x-35\\=x(x-7)+5(x-7)\\=(x-7)(x+5)$

So the zeroes are $2+\sqrt{3}\ ,\ 2-\sqrt{3},7\ and\ -5$

The polynomial division is carried out as follows

Given the remainder =x+a

The obtained remainder after division is $(2k-9)x+10-k(8-k)$

now equating the coefficient of x

$2k-9=1$

which gives the value of $k=5$

now equating the constants

$a=10-k(8-k)=10-5(8-5)=-5$

Therefore k=5 and a=-5

## NCERT solutions for class 10 maths chapter wise

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

## 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 solutions of NCERT class 10 maths chapter 2 polynomials, you must have learnt to answer a question in the step by step manner.

• Once you have studied the NCERT solutions for class 10, your next target should be previous year questions papers.

• NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After going through these two things, you can jump to the next chapters.

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