HSA Review Do Now #1 ☺

Solve these ratio problems:

1. There are 28 students in a class. The ratio of girls : boys is 3:4. How many girls are in the class?

2. Two angles have a sum of 180°. If the measures of the angles have the ratio of 2:3, what is the measure of the smallest angle?

3. Yesterday, 20% of the students in an Algebra class were absent. If 5 students were absent, how many students are in the class?

Introduction to Polynomials

Classifying and

Writing in Standard Form

Definitions

• A monomial is an algebraic expression that is either a constant, a variable, or a product of a constant and one or more variables.
• The constant is called the coefficient .
• A polynomial is a monomial, or a sum or difference of monomials.

Classifying polynomials-Part I

• Mono mial – 1 term
• Bi nomial – 2 terms separated by + or –
• Tri nomial – 3 terms separated by + or –
• Poly nomial – many terms separated by + or -

What kind of polynomial?

• 12x 2 – 5x + 1
• 4x 5 – 10x 4 + 13x 2 – 6x – 9

Classifying polynomials-Part II

• The degree of a polynomial is determined by the exponent with the greatest value within the polynomial.

We can classify a polynomial �by its degree:

If the exponent is greater than 5, follow this:

Classify the polynomial by degree and number of terms.

• The terms of a polynomial may appear in any order. However, in standard form the terms are ordered from left to right in descending order, which mean from the greatest exponent to the least.

Standard Form Examples

Be sure to move the sign as well.

Write in standard form.

## Mathematics Presentation for Class 10

Mathematics, chapter 1: real number, chapter 2: polynomials, chapter 3: pair of linear equation 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.

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## Chapter 2 Class 10 Polynomials

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Get NCERT Solutions of C hapter 2 Class 10 Polynomials free at Teachoo. All NCERT Exercise Questions, Examples and Optional Questions have been solved with video of each and every question.

In this chapter, we will learn

• What is a polynomial
• What are monomial, binomials, trinomials
• What is the degree of polynomial
• What are linear, quadratic and cubic polynomials
• What are Zeroes of a Polynomials

• How to find Zeroes of a Polynomials (both quadratic and cubic)
• Cubic Polynomial
• Dividing two polynomials, and verifying the Division Algorithm for Polynomials

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• NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials
• NCERT Solutions

## NCERT Class 10 Maths Chapter 2: Complete Resource for Polynomials

Mathematics is a crucial subject for Class 10 students. The syllabus is designed in such a way that the students can gather knowledge and develop a conceptual foundation to carry on to the advanced classes. Class 10 Maths Chapter 2 will need special assistance from the NCERT Solutions for Class 10 Maths Chapter 2 prepared by the top mentors of Vedantu. Download the Class 10 Maths Chapter 2 Solutions PDF file for free and access it offline for your convenience.

Chapter 2 Maths Class 10 is based on polynomials. The different types of equations and their components have been described in this NCERT Maths Class 10 Chapter 2. You can easily learn the new concepts and solve the exercise questions by using the NCERT Solutions Class 10 Maths Chapter 2 and complete this chapter. Refer to the solutions while practising the problems and resolve your doubts in no time. NCERT Solutions for all subjects and classes are available on Vedantu. Science students who are looking for NCERT Solutions for Class 10 Science will also find the solutions curated by our Master Teachers really helpful.

## Exercises under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Ncert solutions for class 10 maths chapter 2, "polynomials," is based on the concept of polynomials and their applications. the chapter consists of the following exercises: exercise 2.1: this exercise discusses the concept of polynomials and their terms. exercise 2.2: this exercise covers the addition and subtraction of polynomials. exercise 2.3: this exercise explains the multiplication of polynomials and the use of identities such as (a+b)², (a-b)², and (a+b)(a-b). exercise 2.4: this exercise deals with the factorization of quadratic polynomials., access ncert solutions for class 10 maths chapter 2 – polynomials, exercise - 2.1.

1. The graphs of $\text{y=p(x)}$ are given in following figure, for some Polynomials $\text{p(x)}$. Find the number of zeroes of $\text{p(x)}$, in each case.

Ans: The graph does not intersect the $\text{x-axis}$ at any point. Therefore, it does not have any zeroes.

Ans: The graph intersects at the $\text{x-axis}$ at only $\text{1}$point. Therefore, the number of zeroes is $\text{1}$.

Ans: The graph intersects at the $\text{x-axis}$ at $\text{3}$ points. Therefore, the number of zeroes is $\text{3}$.

Ans: The graph intersects at the $\text{x-axis}$ at $\text{2}$ points. Therefore, the number of zeroes is $\text{2}$.

Ans: The graph intersects at the $\text{x-axis}$ at $\text{4}$ points. Therefore, the number of zeroes is $\text{4}$.

Ans: The graph intersects at the $\text{x-axis}$ at $\text{3}$ points. Therefore, the number of zeroes is $\text{3}$.

## Exercise - 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$

Ans:   Given: ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$.

Now factorize the given polynomial to get the roots.

$\Rightarrow \text{(x}-\text{4)(x+2)}$

The value of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ is zero.

when $\text{x}-\text{4=0}$ or $\text{x+2=0}$. i.e., $\text{x = 4}$ or $\text{x = }-\text{2}$

Therefore, the zeroes of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ are $\text{4}$ and $-2$.

Now, Sum of zeroes$\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ $\therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}}$

Product of zeroes $\text{=4 }\times\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Product of zeroes}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .

(ii) $\mathbf{4{{s}^{2}}-4s+1}$

Ans: Given: $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$

$\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}$

The value of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ is zero.

when $\text{2s}-\text{1=0}$, $\text{2s}-\text{1=0}$. i.e., $\text{s =}\dfrac{\text{1}}{\text{2}}$ and $\text{s =}\dfrac{\text{1}}{\text{2}}$

Therefore, the zeroes of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ are $\dfrac{\text{1}}{\text{2}}$ and $\dfrac{\text{1}}{\text{2}}$.

Now, Sum of zeroes$\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

Product of zeroes$=\dfrac{\text{1}}{\text{2}}\text{ }\times\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

$\therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$ .

(iii) $\mathbf{6{{x}^{2}}-3-7x}$

Ans: Given: $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}$

$\Rightarrow \text{(3x+1)(2x}-\text{3)}$

The value of  $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ is zero.

when $\text{3x+1=0}$ or $\text{2x}-\text{3=0}$. i.e., $\text{x =}\dfrac{-\text{1}}{\text{3}}$ or $\text{x =}\dfrac{\text{3}}{\text{2}}$.

Therefore, the zeroes of  $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ are $\dfrac{\text{-1}}{\text{3}}$ and $\dfrac{\text{3}}{\text{2}}$.

Now, Sum of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

Product of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\times\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

(iv) $\mathbf{4{{u}^{2}}+8u}$

Ans: Given: $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$

$\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}$

$\Rightarrow \text{4u(u+2)}$

The value of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ is zero.

when $\text{4u=0}$ or $\text{u+2=0}$. i.e., $\text{u = 0}$ or $\text{u =}-\text{2}$

Therefore, the zeroes of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ are $\text{0}$ and $\text{-2}$.

Now, Sum of zeroes$\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

Product of zeroes$\text{=0 }\times\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

(v) $\mathbf{{{t}^{2}}-15}$

Ans: Given: ${{\text{t}}^{\text{2}}}-\text{15}$

$\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}$

$\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}$

The value of ${{\text{t}}^{\text{2}}}-\text{15}$ is zero.

when $\text{t}-\sqrt{\text{15}}\text{=0}$ or $\text{t+}\sqrt{\text{15}}\text{=0}$, i.e., $\text{t=}\sqrt{\text{15}}$ or $\text{t=}-\sqrt{\text{15}}$

Therefore, the zeroes of ${{\text{t}}^{\text{2}}}-\text{15}$ are $\sqrt{\text{15}}$ and $-\sqrt{\text{15}}$.

Now, Sum of zeroes$\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

Product of zeroes$\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$ .

(vi) $\mathbf{3{{x}^{2}}-x-4}$

Ans:   Given: $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$

$\Rightarrow \left( 3x-4 \right)(x+1)$

The value of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ is zero.

when $\text{3x}-\text{4=0}$ or $\text{x+1=0}$, i.e., $\text{x=}\dfrac{\text{4}}{\text{3}}$ or $\text{x=}-\text{1}$

Therefore, the zeroes of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ are $\dfrac{\text{4}}{\text{3}}$ and $\text{-1}$.

Now, Sum of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

Product of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i)  $\mathbf{\dfrac{1}{4},-1}$

Ans: Given: $\dfrac{\text{1}}{\text{4}}\text{,-1}$

Let the zeroes of polynomial be $\text{ }\alpha\text{ }$ and $\text{ }\beta\text{ }$.

Then,

$\text{ }\alpha\text{ + }\beta\text{ =}\dfrac{\text{1}}{\text{4}}$

$\text{ }\alpha\text{ }\beta\text{ =}-\text{1}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1$

$\Rightarrow 4{{x}^{2}}-x-4$

Therefore, the quadratic polynomial is $\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$.

(ii) $\mathbf{\sqrt{2},\dfrac{1}{3}}$

Ans: Given: $\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}$

Let the zeroes of polynomial be $\text{ }\alpha \text{ }$ and $\text{ }\beta\text{ }$.

Then, $\text{ }\alpha\text{ + }\beta\text{ =}\sqrt{\text{2}}$

$\text{ }\alpha\text{ }\beta\text{ =}\dfrac{\text{1}}{\text{3}}$

$\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}$

$\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$

Therefore, the quadratic polynomial is $3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$.

(iii) $\mathbf{0,\sqrt{5}}$

(here, root is missing)

Ans: Given: $\text{0,}\sqrt{\text{5}}$

Let the zeroes of polynomial be $\text{ } \alpha \text{ }$ and $\text{ } \beta \text{ }$.

$\text{ } \alpha \text{ + } \beta \text{ = 0}$

$\text{ } \alpha \text{ } \beta \text{ =}\sqrt{\text{5}}$

$\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}$

Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}$.

(iv) $\mathbf{1,1}$

Ans: Given: $\text{1,1}$

$\text{ } \alpha \text{ + } \beta \text{ =1}$

$\text{ } \alpha \text{ } \beta \text{ =1}$

$\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}$

Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}-\text{x+1}$.

(v) $\mathbf{-\dfrac{1}{4},\dfrac{1}{4}}$

Ans: Given: $-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}$

$\text{ } \alpha \text{ + } \beta\text{ =}-\dfrac{\text{1}}{\text{4}}$

$\text{ } \alpha \text{ } \beta \text{ =}\dfrac{\text{1}}{\text{4}}$

$\Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4}$

$\Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$

Therefore, the quadratic polynomial is  $\text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$ .

(vi) $\mathbf{4,1}$

Ans: Given: $\text{4,1}$

Let the zeroes of polynomial be $\text{ } \alpha \text{ }$ and $\text{ }\beta \text{ }$.

$\text{ } \alpha \text{ + } \beta \text{ = 4}$

$\Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1}$

Therefore, the quadratic polynomial is  ${{\text{x}}^{\text{2}}}-\text{4x+1}$ .

## Exercise - 2.3

1. Divide the polynomial $\text{p(x)}$ by the polynomial $\text{g(x)}$ and find the quotient and remainder in each of the following:

(i) $\mathbf{p(x) = {{x}^{3}}-3{{x}^{2}}+5x-3, g(x)={{x}^{2}}-2}$

Ans: Given: $\text{p(x)=}{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+5x}-\text{3}$ and $\text{g(x)=}{{\text{x}}^{\text{2}}}-\text{2}$

Then, divide the polynomial $\text{p(x)}$ by $\text{g(x)}$.

\begin{align} & {{\text{x}}^{\text{2}}}-\text{2}\overset{\text{x}-\text{3}}{\overline{\left){\begin{align} & {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+5x}-\text{3} \\ & {{\text{x}}^{\text{3}}}\text{ }-\text{2x} \\ & \underline{-\text{ + }} \\ \end{align}}\right.}} \\ & \text{ }-\text{3}{{\text{x}}^{\text{2}}}\text{+7x}-\text{3} \\ & \text{ }-\text{3}{{\text{x}}^{\text{2}}}\text{ + 6} \\ & \text{ }\underline{\text{+ }-\text{ }} \\ & \text{ }\underline{\text{ 7x}-\text{9 }} \\ & \text{ } \\ \end{align}

Therefore, Quotient$\text{= x}-\text{3}$and Remainder$\text{= 7x}-\text{9}$.

(ii) $\mathbf{p(x) = {{x}^{4}}-3{{x}^{2}}+ 4x + 5, g(x) = {{x}^{2}}+1-x}$

Ans: Given: $\text{p(x) = }{{\text{x}}^{\text{4}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5}\Rightarrow {{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5}$

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

\begin{align} & {{\text{x}}^{\text{2}}}-\text{x+1}\overset{{{\text{x}}^{\text{2}}}\text{+x}-\text{3}}{\overline{\left){\begin{align} & {{\text{x}}^{\text{4}}}\text{ + 0}{{\text{x}}^{3}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 4x + 5} \\ & {{\text{x}}^{\text{4}}} -{{\text{x}}^{\text{3}}}\text{ + }{{\text{x}}^{\text{2}}} \\ \end{align}}\right.}} \\ & \text{ }\underline{-\text{ + }-\text{ }} \\ & \text{ }{{\text{x}}^{\text{3}}} -\text{4}{{\text{x}}^{\text{2}}}\text{+ 4x +5} \\ & \text{ }{{\text{x}}^{\text{3}}} -{{\text{x}}^{\text{2}}}\text{ +x} \\ & \text{ }\underline{\text{ }-\text{ + }-\text{ }} \\ & \text{ }-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x + 5} \\ & \text{ }\underline{\begin{align} & \text{ }-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x}-\text{3} \\ & \text{ + }-\text{ + } \\ \end{align}} \\ & \text{ }\underline{\text{ 8 }} \\ \end{align}

Therefore, Quotient$\text{= }{{\text{x}}^{\text{2}}}\text{+ x}-\text{3}$ and Remainder$\text{=8}$.

(iii)  $\mathbf{p(x)={{x}^{4}}-5x+6, g(x)=2-{{x}^{2}}}$

Ans: Given: $\text{p(x) = }{{\text{x}}^{\text{4}}}-\text{5x+6}\Rightarrow {{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{3}}\text{+0}{{\text{x}}^{\text{2}}}-\text{5x+6}$

$\text{g(x) = 2}-{{\text{x}}^{\text{2}}}\Rightarrow \text{ }-{{\text{x}}^{\text{2}}}\text{+2}$

\begin{align} & -{{\text{x}}^{\text{2}}}\text{+2}\overset{-{{\text{x}}^{\text{2}}}-\text{2}}{\overline{\left){{{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{3}}\text{+ 0}{{\text{x}}^{\text{2}}}-\text{5x+6}}\right.}} \\ & \text{ }{{\text{x}}^{\text{4}}}\text{ }-\text{2}{{\text{x}}^{\text{2}}} \\ & \text{ }\underline{-\text{ + }} \\ & \text{ 2}{{\text{x}}^{\text{2}}}-\text{5x + 6} \\ & \text{ 2}{{\text{x}}^{\text{2}}}\text{ }-\text{4} \\ & \text{ }\underline{\text{ }-\text{ + }} \\ & \text{ }\underline{\text{ }-\text{5x + 10}} \\ \end{align}

Therefore, Quotient$\text{=}-{{\text{x}}^{\text{2}}}-\text{2}$ and Remainder$\text{=}-\text{5x+10}$

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i)  $\mathbf{{{t}^{2}}-3, 2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12}$

Ans: Given: ${{\text{t}}^{\text{2}}}-\text{3, 2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}$

Let us take first polynomial is  ${{\text{t}}^{\text{2}}}-\text{3}\Rightarrow \text{ }{{\text{t}}^{\text{2}}}\text{+0t}-\text{3}$ .

And second polynomial is $\text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}$.

Now, divide the second polynomial by first polynomial.

{{\text{t}}^{\text{2}}}\text{+ 0}{{\text{t}}^{\text{2}}}-\text{3}\overset{\text{2}{{\text{t}}^{\text{2}}}\text{+ 3t + 4}}{\overline{\left){\begin{align} & \text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12} \\ & \text{2}{{\text{t}}^{\text{4}}}\text{+0}{{\text{t}}^{\text{3}}}-\text{6}{{\text{t}}^{\text{2}}} \\ & \underline{-\text{ }-\text{ + }} \\ & \text{ 2}{{\text{t}}^{\text{3}}}\text{+ 4}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12} \\ & \text{ 3}{{\text{t}}^{\text{3}}}\text{+ 0}{{\text{t}}^{\text{2}}}-\text{9t} \\ & \underline{\text{ }-\text{ }-\text{ + }} \\ & \text{ 4}{{\text{t}}^{\text{2}}}\text{+0t}-\text{12} \\ & \underline{\text{ 4}{{\text{t}}^{\text{2}}}\text{+0t}-\text{12 }} \\ & \underline{\text{ 0 }} \\ \end{align}}\right.}}

Since the remainder is $\text{0}$

Therefore, ${{\text{t}}^{\text{2}}}-\text{3}$ is a factor of $\text{2}{{\text{t}}^{\text{4}}}\text{+3}{{\text{t}}^{\text{3}}}-\text{2}{{\text{t}}^{\text{2}}}-\text{9t}-\text{12}$.

(ii) $\mathbf{{{x}^{2}}+3x+1, 3{{x}^{4}}+5{{x}^{3}}-7{{x}^{2}}+2x+2}$

Ans: Given: ${{\text{x}}^{\text{2}}}\text{+ 3x + 1, 3}{{\text{x}}^{\text{4}}}\text{+5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x+ 2}$

Let us take first polynomial is ${{\text{x}}^{\text{2}}}\text{+ 3x + 1}$.

And second polynomial is $\text{3}{{\text{x}}^{\text{4}}}\text{+5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x+ 2}$

{{\text{x}}^{\text{2}}}\text{+ 3x + 1}\overset{\text{3}{{\text{x}}^{\text{2}}}-\text{ 4x + 2}}{\overline{\left){\begin{align} & \text{3}{{\text{x}}^{\text{4}}}\text{+ 5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x + 2} \\ & \text{3}{{\text{x}}^{\text{4}}}\text{+ 9}{{\text{x}}^{\text{3}}}\text{+ 3}{{\text{x}}^{\text{2}}} \\ & \underline{-\text{ }-\text{ }-\text{ }} \\ & \text{ }-\text{4}{{\text{x}}^{\text{3}}}-\text{10}{{\text{x}}^{\text{2}}}\text{+ 2x + 2} \\ & \text{ }-\text{4}{{\text{x}}^{\text{3}}}-\text{12}{{\text{x}}^{\text{2}}}-\text{4x} \\ & \underline{\text{ + + + }} \\ & \text{ 2}{{\text{x}}^{\text{2}}}\text{+ 6x + 2} \\ & \text{ 2}{{\text{x}}^{\text{2}}}\text{+ 6x + 2} \\ & \underline{\text{ }-\text{ }-\text{ }-\text{ }} \\ & \underline{\text{ 0 }} \\ \end{align}}\right.}}

Therefore, ${{\text{x}}^{\text{2}}}\text{+ 3x + 1}$ is a factor of $\text{3}{{\text{x}}^{\text{4}}}\text{+ 5}{{\text{x}}^{\text{3}}}-\text{7}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}$.

(iii) $\mathbf{{{x}^{2}}-3x+1, {{x}^{5}}-4{{x}^{3}}+{{x}^{2}}+3x+1}$

Ans: Given: ${{\text{x}}^{\text{2}}}-\text{3x + 1, }{{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1}$.

Let us take first polynomial is ${{\text{x}}^{\text{2}}}-\text{3x + 1}$.

And second polynomial is ${{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1}$.

{{\text{x}}^{\text{2}}}-\text{3x + 1}\overset{{{\text{x}}^{\text{2}}}-\text{1}}{\overline{\left){\begin{align} & {{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x+1} \\ & {{\text{x}}^{\text{5}}}-\text{3}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}} \\ & \underline{-\text{ + }-\text{ }} \\ & \text{ }-{{\text{x}}^{\text{3}}}\text{ + 3x +1} \\ & \text{ }-{{\text{x}}^{\text{3}}}\text{ + 3x}-\text{1} \\ & \underline{\text{ + }-\text{ + }} \\ & \text{ }\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,} \end{align}}\right.}}

Since the remainder$\ne \text{0}$.

Therefore, ${{\text{x}}^{\text{2}}}-\text{3x + 1}$ is not a factor of ${{\text{x}}^{\text{5}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+ }{{\text{x}}^{\text{2}}}\text{+ 3x +1}$.

3.Obtain all other zeroes of $\text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5}$, if two of its zeroes are $\sqrt{\dfrac{\text{5}}{\text{3}}}$ and $-\sqrt{\dfrac{\text{5}}{\text{3}}}$

Ans: Let us assume, $\text{p(x)=3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}\text{-2}{{\text{x}}^{\text{2}}}\text{-10x-5}$

Then, given two zeroes are $\sqrt{\dfrac{\text{5}}{\text{3}}}$ and $-\sqrt{\dfrac{\text{5}}{\text{3}}}$.

$\therefore \left( \text{x-}\sqrt{\dfrac{\text{5}}{\text{3}}} \right)\left( \text{x+}\sqrt{\dfrac{\text{5}}{\text{3}}} \right)\text{=}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)$ is a factor of  $\text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5}$

Now, divide the given polynomial by ${{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}}$

{{\text{x}}^{\text{2}}}\text{+ 0x}-\dfrac{\text{5}}{\text{3}}\overset{\text{3}{{\text{x}}^{\text{2}}}\text{+ 6x + 3}}{\overline{\left){\begin{align} & \text{3}{{\text{x}}^{\text{4}}}\text{+6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5} \\ & \text{3}{{\text{x}}^{\text{4}}}\text{+0}{{\text{x}}^{\text{3}}}-\text{5}{{\text{x}}^{\text{2}}} \\ & \underline{-\text{ }-\text{ + }} \\ & \text{ 6}{{\text{x}}^{\text{3}}}\text{+ 3}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5} \\ & \text{ 6}{{\text{x}}^{\text{3}}}\text{+ 0}{{\text{x}}^{\text{2}}}-\text{10x} \\ & \underline{ -\text{ }-\text{ + }} \\ & \text{ 3}{{\text{x}}^{\text{2}}}\text{+ 0x}-\text{5} \\ & \text{ 3}{{\text{x}}^{\text{2}}}\text{+ 0x}-\text{5} \\ & \underline{\text{ }-\text{ }-\text{ + }} \\ & \underline{\text{ 0 }} \\ & \text{ } \\ \end{align}}\right.}}

$\therefore \text{3}{{\text{x}}^{\text{4}}}\text{+ 6}{{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{10x}-\text{5=}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)\left( \text{3}{{\text{x}}^{\text{2}}}\text{+6x+3} \right)$

$\Rightarrow \text{3}\left( {{\text{x}}^{\text{2}}}-\dfrac{\text{5}}{\text{3}} \right)\left( {{\text{x}}^{\text{2}}}\text{+ 2x + 1} \right)$

$\Rightarrow \left( \text{3}{{\text{x}}^{\text{2}}}-\text{5} \right)\left( {{\text{x}}^{\text{2}}}\text{+ 2x +1} \right)$

Now, factorize the polynomial ${{\text{x}}^{\text{2}}}\text{+ 2x +1}$

$\Rightarrow {{\left( \text{x+1} \right)}^{\text{2}}}$

Hence, its zero is given by $\text{x+1=0}$

$\Rightarrow \text{x=}-\text{1}$

As it has the term ${{\left( \text{x+1} \right)}^{\text{2}}}$, then, there will be $\text{2}$ zeroes at $\text{x =}-\text{1}$.

Therefore, the zeroes of the given polynomial are $\sqrt{\dfrac{\text{5}}{\text{3}}}\text{,}-\sqrt{\dfrac{\text{5}}{\text{3}}}\text{,}-\text{1}$ and $-\text{1}$.

4. On dividing ${{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2}$ by a polynomial $\text{g(x)}$, the quotient and remainder were $\text{x}-\text{2}$ and $-\text{2x + 4}$, respectively. Find $\text{g(x)}$.

Ans: Let us take the dividend as $\text{p(x) }$. Then, $\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2}$

And the divisor is $\text{g(x)}$. Then, find the value of $\text{g(x)}$.

Quotient$\text{=}\left( \text{x}-\text{2} \right)$

Remainder$\text{=(}-\text{2x+ 4)}$

$\text{Dividend = Divisor }\times \text{ Quotient + Remainder}$

${{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2 = g(x) } \times \text{ (x}-\text{2)+(}-\text{2x+4)}$

${{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x + 2 + 2x}-\text{4 = g(x)}\times \text{(x}-\text{2)}$

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

Hence, $\text{g(x)}$ is the quotient when we divide $\left( {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+3x}-\text{2} \right)$ by $\left( \text{x}-\text{2} \right)$.

\text{x}-\text{2}\overset{{{\text{x}}^{\text{2}}}-\text{x+1}}{\overline{\left){\begin{align} & {{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ 3x }-\text{ 2} \\ & {{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}} \\ & \underline{-\text{ + }} \\ & \text{ }-{{\text{x}}^{\text{2}}}\text{+ 3x}-\text{2} \\ & \text{ }-{{\text{x}}^{\text{2}}}\text{+ 2x} \\ & \underline{\text{ + }-\text{ }} \\ & \underline{\begin{align} & \text{ x}-2 \\ & \text{ x}-2 \\ & \underline{\text{ }-\text{ + }} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{align}} \\ \end{align}}\right.}}

$\therefore \text{g(x)=}\left( {{\text{x}}^{\text{2}}}-\text{x+1} \right)$

5. Give examples of polynomial $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and

(i) $\mathbf{deg p(x)=deg q(x)}$

Ans: According to the division algorithm, if $\text{p(x)}$ and $\text{g(x)}$ are two polynomials with

$\text{g(x)}\ne \text{0}$, then we can find polynomials $\text{q(x)}$ and $\text{r(x)}$ such that

$\text{p(x) = g(x) }\times \text{ q(x) + r(x)}$,

Where $\text{r(x)=0}$ or degree of $\text{r(x)}$ degree of $\text{g(x)}$

Degree of a polynomial is the highest power of the variable in the polynomial.

Given: $\text{deg p(x) = deg q(x)}$

The degree of the quotient will be equal to the degree of dividend when the divisor is constant (i.e., when any polynomial is divided by a constant).

Let us assume the division of $\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}$ by $\text{2}$.

Here, $\text{p(x) = 6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}$

$\text{g(x) = 2}$

$\text{q(x) = 3}{{\text{x}}^{\text{2}}}\text{+ x + 1}$ and $\text{r(x) = 0}$

Degree of $\text{p(x)}$and $\text{q(x)}$ is the same i.e., $\text{2}$.

Checking for division algorithm,

$\text{p(x) = g(x) }\times\text{ q(x) + r(x)}$

$\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2 = 2}\left( \text{3}{{\text{x}}^{\text{2}}}\text{+ x + 1} \right)$

$\text{6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}\,\text{= 6}{{\text{x}}^{\text{2}}}\text{+ 2x + 2}$

Therefore, the division algorithm is satisfied.

(ii) $\mathbf{deg q(x)=deg r(x)}$

Ans: Given: $\text{deg q(x) = deg r(x)}$

Let us assume the division of ${{\text{x}}^{\text{3}}}\text{+x}$ by ${{\text{x}}^{\text{2}}}$

Here, $\text{p(x) = }{{\text{x}}^{\text{3}}}\text{+ x}$

$\text{g(x) = }{{\text{x}}^{\text{2}}}$

$\text{q(x) = x}$ and $\text{r(x) = x}$

Clearly, the degree of $\text{q(x)}$ and $\text{r(x)}$ is the same, i.e.,

${{\text{x}}^{\text{3}}}\text{+ x =}\left( {{\text{x}}^{\text{2}}} \right)\text{ }\times\text{ x + x}$

${{\text{x}}^{\text{3}}}\text{+ x = }{{\text{x}}^{\text{3}}}\text{+ x}$

Therefore, the division algorithm is satisfied.

(iii) $\mathbf{deg r\left( x \right)=0 }$

Ans: Given: $\text{deg r(x) = 0}$

Degree of remainder will be $\text{0}$ when remainder comes to a constant.

Let us assume the division of ${{\text{x}}^{\text{3}}}\text{+1}$ by ${{\text{x}}^{\text{2}}}$.

Here, $\text{p(x) = }{{\text{x}}^{\text{3}}}\text{+1}$

$\text{q(x) = x}$ and $\text{r(x) =1}$

Clearly, the degree of $\text{r(x)}$ is $\text{0}$

${{\text{x}}^{\text{3}}}\text{+1=}\left( {{\text{x}}^{\text{2}}} \right)\text{ }\times\text{ x +1}$

${{\text{x}}^{\text{3}}}\text{+1= }{{\text{x}}^{\text{3}}}\text{+1}$

## Exercise - 2.4

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:

(i) $\mathbf{2{{x}^{3}}+ {{x}^{2}}-5x + 2; \dfrac{1}{2},1,-2}$

Ans: Let us assume $\text{p(x) = 2}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}-\text{5x + 2}$

And zeroes for this polynomial are $\dfrac{\text{1}}{\text{2}}\text{,1,}-\text{2}$. Then,

$\text{p}\left( \dfrac{\text{1}}{\text{2}} \right)\text{ = 2}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{3}}}\text{+}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}-\text{5}\left( \dfrac{\text{1}}{\text{2}} \right)\text{+ 2}$

$\text{=}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{4}}-\dfrac{\text{5}}{\text{2}}\text{+2}$

$\text{= 0}$

$\text{p(1) = 2 }\times \text{ }{{\text{1}}^{\text{3}}}\text{+}{{\text{1}}^{2}}-\text{5 }\times\text{ 1+2}$

$\text{= 0}$

$\text{p(}-\text{2)=2(}-\text{2}{{\text{)}}^{\text{3}}}\text{+}{{\left( -\text{2} \right)}^{\text{2}}}-\text{5}\left( -\text{2} \right)\text{+2}$

$\text{=}-\text{16 + 4 +10 + 2}$

Therefore, $\dfrac{\text{1}}{\text{2}}\text{,1}$ and $\text{-2}$ are the zeroes of the given polynomial.

Comparing the given polynomial with $\text{a}{{\text{x}}^{\text{3}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+cx+d}$ to get, $\text{a = 2, b = 1, c =}-\text{5,}$ and $\text{d = 2}$.

Let us take $\text{ } \alpha\text{ =}\dfrac{\text{1}}{\text{2}}\text{, }\beta\text{ =1,}$ and $\text{ }\gamma\text{ =}-\text{2}$.

$\text{ } \alpha \text{ + }\beta\text{ + } \gamma\text{ =}\dfrac{\text{1}}{\text{2}}\text{+1+}\left( -\text{2} \right)\text{=}-\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{-\text{b}}{\text{a}}$

$\therefore \text{ } \alpha \text{ + } \beta \text{ + } \gamma\text{ =}\dfrac{-\text{b}}{\text{a}}$

$\text{ } \alpha \text{ } \beta \text{ + } \beta \text{ } \gamma \text{ + } \alpha \text{ } \gamma \text{ = }\dfrac{\text{1}}{\text{2}}\text{ } \times \text{ 1+1}\left( -\text{2} \right)\text{+}\dfrac{\text{1}}{\text{2}}\left( -\text{2} \right)\text{ = }\dfrac{-\text{5}}{\text{2}}\text{ = }\dfrac{\text{c}}{\text{a}}$

$\therefore \text{ } \alpha \text{ }\beta\text{ + }\beta \text{ }\gamma\text{ + } \alpha\text{ } \gamma \text{ =}\dfrac{\text{c}}{\text{a}}$

$\text{ }\alpha\text{ }\beta\text{ }\gamma \text{ =}\dfrac{\text{1}}{\text{2}}\text{ } \times\text{ 1 }\times \text{ }\left( -\text{2} \right)\text{ = }\dfrac{-\text{1}}{\text{1}}\text{ = }\dfrac{-\left( \text{2} \right)}{\text{2}}\text{ = }\dfrac{-\text{d}}{\text{a}}$

$\therefore \text{ } \alpha\text{ } \beta \text{ } \gamma \text{ = }\dfrac{-\text{d}}{\text{a}}$

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) $\mathbf{{{x}^{3}}-4{{x}^{2}}+5x-2; 2,1,1}$

Ans: Let us assume $\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{4}{{\text{x}}^{\text{2}}}\text{+ 5x}-\text{2}$

And zeroes for this polynomial are $\text{2,1,1}$. Then,

$\text{p(2) = }{{\text{2}}^{\text{3}}}-\text{4}\left( {{\text{2}}^{\text{2}}} \right)\text{+5}\left( \text{2} \right)-\text{2}$

$\text{= 8}-\text{16+10}-\text{2}$

$\text{= 0}$

$\text{p(1) = }{{\text{1}}^{\text{3}}}-\text{4}\left( {{\text{1}}^{\text{2}}} \right)\text{+5}\left( \text{1} \right)-\text{2}$

$\text{=1}-\text{4+5}-\text{2}$

$\text{= 0}$

Therefore, $\text{2,1,}$ and $\text{1}$ are the zeroes of the given polynomial.

Comparing the given polynomial with $\text{a}{{\text{x}}^{\text{3}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+cx+d}$ to get,$\text{a = 1, b =}-\text{4, c = 5,}$and $\text{d =}-\text{2}$.

Let us take $\text{ }\alpha\text{ = 2, }\beta \text{ =1,}$ and $\text{ } \gamma \text{ =1}$.

$\text{ } \alpha \text{ + } \beta \text{ + } \gamma \text{ = 2+1+1= 4 =}\dfrac{-\text{(}-\text{4)}}{\text{1}}=\dfrac{-\text{b}}{\text{a}}$

$\therefore \text{ } \alpha \text{ + } \beta \text{ + } \gamma \text{ =}\dfrac{-\text{b}}{\text{a}}$

$\text{ } \alpha \text{ } \beta \text{ + } \beta \text{ } \gamma \text{ + } \alpha \text{ } \gamma \text{ =(2)(1)+(1)(1)+(2)(1)}$

$\text{= 2+1+2 = 5 =}\dfrac{\left( \text{5} \right)}{\text{1}}\text{=}\dfrac{\text{c}}{\text{a}}$

$\therefore \text{ } \alpha \text{ } \beta \text{ + } \beta \text{ } \gamma \text{ + } \alpha \text{ } \gamma \text{ =}\dfrac{\text{c}}{\text{a}}$

$\text{ } \alpha \text{ } \beta \text{ } \gamma \text{ = 2 } \times \text{ 1 } \times \text{ 1= 2 =}\dfrac{-\left( -\text{2} \right)}{\text{1}}\text{=}\dfrac{-\text{d}}{\text{a}}$

$\therefore \text{ } \alpha \text{ } \beta \text{ }\gamma\text{ = }\dfrac{-\text{d}}{\text{a}}$

2. 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 $\text{2,}-\text{7,}-\text{14}$ respectively.

Ans: Let the polynomial be $\text{a}{{\text{x}}^{\text{3}}}\text{+ b}{{\text{x}}^{\text{2}}}\text{+ cx + d}$ and the zeroes be $\text{ }\alpha\text{ , }\beta\text{ ,}$ and $\text{ }\gamma\text{ }$.

Then given that,

$\text{ }\alpha\text{ + }\beta\text{ + }\gamma\text{ =}\dfrac{\text{2}}{\text{1}}\text{=}\dfrac{-\text{b}}{\text{a}}$

$\text{ }\alpha\text{ }\beta\text{ + }\beta\text{ }\gamma\text{ + }\alpha\text{ }\gamma\text{ =}\dfrac{-\text{7}}{\text{1}}\text{=}\dfrac{\text{c}}{\text{a}}$

$\text{ }\alpha\text{ }\beta\text{ }\gamma\text{ =}\dfrac{-\text{14}}{\text{1}}\text{=}\dfrac{-\text{d}}{\text{a}}$

If $\text{a=1}$, then $\text{b =}-\text{2, c =}-\text{7, d =14}$

Therefore, the polynomial is ${{\text{x}}^{\text{3}}}-\text{2}{{\text{x}}^{\text{2}}}-\text{7x +14}$.

3. If the zeroes of polynomial ${{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x +1}$ are $\text{a}-\text{b, a, a+b}$, find $\text{a}$ and $\text{b}$.

Ans: Let us assume $\text{p(x) = }{{\text{x}}^{\text{3}}}-\text{3}{{\text{x}}^{\text{2}}}\text{+ x +1}$.

And the zeroes are $\text{a-b, a, a+b}$.

Let us assume  $\text{ } \alpha\text{ = a-b, }\beta\text{ = a}$  and  $\text{ }\gamma\text{ = a+b}$.

Comparing the given polynomial with $\text{p}{{\text{x}}^{\text{3}}}\text{+ q}{{\text{x}}^{\text{2}}}\text{+ rx + t}$ to get,

$\text{p = 1, q =}-\text{3, r =1,}$ and $\text{t =1}$.

$\text{ }\alpha \text{ + }\beta\text{ + }\gamma\text{ = a}-\text{b+ a + a+b}$

$\Rightarrow \dfrac{-\text{q}}{\text{p}}\text{=3a}$

$\Rightarrow \dfrac{-\left( -\text{3} \right)}{\text{1}}\text{=3a}$

$\Rightarrow \text{3=3a}$

$\therefore \text{a =1}$

Then, the zeroes are $\text{1-b,1+b}$.

$\text{ } \alpha \text{ } \beta \text{ } \gamma \text{ = 1}\left( \text{1-b} \right)\left( \text{1+b} \right)$

$\Rightarrow \dfrac{-\text{t}}{\text{p}}\text{=1}-{{\text{b}}^{\text{2}}}$

$\Rightarrow \dfrac{-\text{1}}{\text{1}}\text{=1}-{{\text{b}}^{\text{2}}}$

$\Rightarrow \text{1}-{{\text{b}}^{\text{2}}}\text{=}-\text{1}$

$\Rightarrow {{\text{b}}^{\text{2}}}=2$

$\therefore \text{b = }\!\!\pm\!\!\text{ }\sqrt{\text{2}}$

Therefore, $\text{a=1}$ and $\text{b =}\pm \sqrt{\text{2}}$.

4. If two zeroes of the polynomial ${{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35}$ are $\text{2 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}$, find other zeroes.

Ans: Given that $\text{2+}\sqrt{\text{3}}$ and $\text{2}-\sqrt{\text{3}}$ are zeroes of the given polynomial.

Therefore, $\left( \text{x}-\text{2}-\sqrt{\text{3}} \right)\left( \text{x}-\text{2+}\sqrt{\text{3}} \right)\text{= }{{\text{x}}^{\text{2}}}\text{+4}-\text{4x}-\text{3}$

$\text{= }{{\text{x}}^{\text{2}}}-\text{4x+1}$

Hence, ${{\text{x}}^{\text{2}}}-\text{4x+1}$ is a factor of the given polynomial.

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing the polynomial ${{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35}$ by ${{\text{x}}^{\text{2}}}-\text{4x+1}$.

{{\text{x}}^{\text{2}}}-\text{4x+1}\overset{{{\text{x}}^{\text{2}}}-\text{2x}-\text{35}}{\overline{\left){\begin{align} & {{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35} \\ & {{\text{x}}^{\text{4}}}-\text{4}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}} \\ & \underline{-\text{ + }-\text{ }} \\ & \text{ }-\text{2}{{\text{x}}^{\text{3}}}-\text{27}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35} \\ & \text{ }-\text{2}{{\text{x}}^{\text{3}}}\text{+ 8}{{\text{x}}^{\text{2}}} -\text{2x} \\ & \underline{\text{ + }-\text{ + }} \\ & \text{ }-\text{35}{{\text{x}}^{\text{2}}}\text{+140x}-\text{35} \\ & \text{ }-\text{35}{{\text{x}}^{\text{2}}}\text{+140x}-\text{35} \\ & \underline{\text{ + }-\text{ + }} \\ & \underline{\text{ 0 }} \\ \end{align}}\right.}}

Clearly, ${{\text{x}}^{\text{4}}}-\text{6}{{\text{x}}^{\text{3}}}-\text{26}{{\text{x}}^{\text{2}}}\text{+138x}-\text{35=}\left( {{\text{x}}^{\text{2}}}-\text{4x+1} \right)\left( {{\text{x}}^{\text{2}}}-\text{2x}-\text{35} \right)$

Then, ${{\text{x}}^{\text{2}}}-\text{2x}-\text{35}$ is also a factor of the given polynomial.

And,${{\text{x}}^{\text{2}}}-\text{2x}-\text{35=}\left( \text{x}-\text{7} \right)\left( \text{x+5} \right)$

Therefore, the value of the polynomial is also zero when $\text{x-7=0}$ Or $\text{x+5=0}$

Hence, $\text{x=7}$ or $\text{-5}$

Therefore, $\text{7}$ and $\text{-5}$ are also zeroes of this polynomial.

5. If the polynomial ${{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10}$ is divided by another polynomial ${{\text{x}}^{\text{2}}}\text{-2x+k}$, the remainder comes out to be $\text{x+a}$, find $\text{k}$ and $\text{a}$.

Ans: Given:${{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10}$ and ${{\text{x}}^{\text{2}}}\text{-2x+k}$.

Then, the remainder is $\text{x+a}$

By division algorithm,

$\text{Dividend = Divisor }\!\!\times\!\!\text{ Quotient+Remainder}$

$\text{Dividend-Remainder= Divisor }\!\!\times\!\!\text{ Quotient}$

${{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-25x-10-x-a}\Rightarrow {{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a}$ will be perfectly divisible by ${{\text{x}}^{\text{2}}}\text{-2x+k}$.

Let us divide ${{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a}$ by ${{\text{x}}^{\text{2}}}\text{-2x+k}$

{{\text{x}}^{\text{2}}}\text{-2x+k}\overset{{{\text{x}}^{\text{2}}}\text{-4x+}\left( \text{8-k} \right)}{\overline{\left){\begin{align} & {{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+16}{{\text{x}}^{\text{2}}}\text{-26x+10-a} \\ & {{\text{x}}^{\text{4}}}\text{-2}{{\text{x}}^{\text{3}}}\text{+k}{{\text{x}}^{\text{2}}} \\ & \underline{\text{- + - }} \\ & \text{ -4}{{\text{x}}^{\text{3}}}\text{+}\left( \text{16-k} \right){{\text{x}}^{\text{2}}}\text{-26x} \\ & \text{ -4}{{\text{x}}^{\text{3}}}\text{+ 8}{{\text{x}}^{\text{2}}}\text{-4kx} \\ & \underline{\text{ + - + }} \\ & \left( \text{8-k} \right){{\text{x}}^{\text{2}}}\text{-}\left( \text{26-4k} \right)\text{x+10 - a} \\ & \left( \text{8-k} \right){{\text{x}}^{\text{2}}}\text{-}\left( \text{16-2k} \right)\text{x+}\left( \text{8k-}{{\text{k}}^{\text{2}}} \right) \\ & \underline{\text{- + - }} \\ & \underline{\left( \text{-10+2k} \right)\text{x+}\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{ }}\text{ } \end{align}}\right.}}

Hence, the reminder $\left( \text{-10+2k} \right)\text{x+}\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)$ will be $\text{0}$.

Then, $\left( \text{-10+2k} \right)\text{=0}$ and $\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{=0}$

For $\left( \text{-10+2k} \right)\text{=0}$

$\text{2k=10}$

$\therefore \text{k=5}$

For $\left( \text{10-a-8k+}{{\text{k}}^{\text{2}}} \right)\text{=0}$

$\text{10-a-8 }\!\!\times\!\!\text{ 5+25=0}$

$\text{10-a-40+25=0}$

$\text{-5-a=0}$

$\therefore \text{a=-5}$

Hence, $\text{k=5}$ and $\text{a=-5}$.

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

You can opt for Chapter 2 - Polynomials NCERT Solutions for Class 10 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

## Important Topics under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Chapter 2 Polynomials is an important chapter in the mathematics syllabus for Class 10. This chapter on Polynomials includes a number of topics, and in order to internalize the ideas properly, students are advised to go through the important topics under Polynomials, thoroughly. We have provided the following list of important topics covered in this chapter for a better understanding of the concept of Polynomials.

Degree of a polynomial

Types of polynomial

Constant polynomial

Linear polynomial

Cubic polynomial

Value of a polynomial

Zero of a polynomial

Graph of a polynomial

Meaning of the zeroes of a quadratic polynomial

## Importance of Polynomials

Polynomials are expressions that have more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers.

They are important because they have applications in most mathematical expressions. They are used to represent appropriate relations between different variables or numbers. We encourage students to learn from this chapter to be able to solve tricky problems easily in exams.

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Polynomials: ncert solutions for class 10 maths chapter 2 summary.

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## NCERT Exemplar for Class 10 Maths

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5. What is a polynomial in Class 10 NCERT?

Ans: Polynomial is an algebraic expression that has variables and coefficients in it. The meaning of a polynomial is “many terms”. An example of a polynomial is $4x^2 + 3x + 7$. They are usually a sum or difference of exponents and variables. For an expression to be a polynomial, it should not contain the square root, negative powers, or fractional powers on the variables. It should also not have variables in the denominator of any fractions.

## NCERT Solutions for Class 10 Maths

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## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Ncert solutions class 10 maths chapter 2 – cbse free pdf download.

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## Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 2 Polynomials

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NCERT Solutions for Class 10 Maths is an extremely important study resource for students. Solving these Polynomials NCERT Solutions of Class 10 Maths would help the students fetch good marks in the board exams. Moreover, experts have focused on following the updated CBSE Syllabus for 2023-24 while preparing these solutions.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• 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
• Exercise 2.1 Chapter 2 Polynomials
• Exercise 2.2 Chapter 2 Polynomials
• Exercise 2.3 Chapter 2 Polynomials
• Exercise 2.4 Chapter 2 Polynomials

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## Access Answers to NCERT Class 10 Maths Chapter 2 – Polynomials

Exercise 2.1 page: 28.

1. 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.

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.

(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.

(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.

(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.

(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.

## Exercise 2.2 Page: 33

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x 2 –2x –8

⇒ x 2 – 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x 2 –2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x 2 )

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x 2 )

(ii) 4s 2 –4s+1

⇒4s 2 –2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s 2 –4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s 2 )

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s 2 )

(iii) 6x 2 –3–7x

⇒6x 2 –7x–3 = 6x 2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x 2 –3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2 )

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x 2 )

(iv) 4u 2 +8u

Therefore, zeroes of polynomial equation 4u 2 + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u 2 )

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u 2 )

(v) t 2 –15

⇒ t 2 = 15 or t = ±√15

Therefore, zeroes of polynomial equation t 2 –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t 2 )

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t 2 )

(vi) 3x 2 –x–4

⇒ 3x 2 –4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x 2 – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x 2 )

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x 2 )

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.

(i) 1/4 , -1

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x 2 –(α+β)x +αβ = 0

x 2 –(1/4)x +(-1) = 0

4x 2 –x-4 = 0

Thus , 4x 2 –x–4 is the quadratic polynomial.

(ii) √2, 1/3

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

x 2  –(√2)x + (1/3) = 0

3x 2 -3√2x+1 = 0

Thus, 3x 2 -3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

x 2 –(0)x +√5= 0

Thus, x 2 +√5 is the quadratic polynomial.

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

x 2 –x+1 = 0

Thus, x 2 –x+1 is the quadratic polynomial.

(v) -1/4, 1/4

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

x 2 –(-1/4)x +(1/4) = 0

4x 2 +x+1 = 0

Thus, 4x 2 +x+1 is the quadratic polynomial.

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

x 2 –(α+β)x+αβ = 0

x 2 –4x+1 = 0

Thus, x 2 –4x+1 is the quadratic polynomial.

## Exercise 2.3 Page: 36

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

Dividend = p(x) = x 3 -3x 2 +5x–3

Divisor = g(x) = x 2 – 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x 4 -3x 2 +4x+5 , g(x) = x 2 +1-x

Dividend = p(x) = x 4 – 3x 2 + 4x +5

Divisor = g(x) = x 2 +1-x

Quotient = x 2 + x–3

Remainder = 8

(iii) p(x) =x 4 –5x+6, g(x) = 2–x 2

Dividend = p(x) =x 4 – 5x + 6 = x 4 +0x 2 –5x+6

Divisor = g(x) = 2–x 2 = –x 2 +2

Quotient = -x 2 -2

Remainder = -5x + 10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t 2 -3, 2t 4 +3t 3 -2t 2 -9t-12

First polynomial = t 2 -3

Second polynomial = 2t 4 +3t 3 -2t 2 -9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t 2 -3 is a factor of 2t 4 +3t 3 -2t 2 -9t-12.

(ii)x 2 +3x+1 , 3x 4 +5x 3 -7x 2 +2x+2

First polynomial = x 2 +3x+1

Second polynomial = 3x 4 +5x 3 -7x 2 +2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x 2 + 3x + 1 is a factor of 3x 4 +5x 3 -7x 2 +2x+2.

(iii) x 3 -3x+1, x 5 -4x 3 +x 2 +3x+1

First polynomial = x 3 -3x+1

Second polynomial = x 5 -4x 3 +x 2 +3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x 3 -3x+1 is not a factor of x 5 -4x 3 +x 2 +3x+1 .

3. Obtain all other zeroes of 3x 4 +6x 3 -2x 2 -10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

∴ (x – √(5/3) ) (x+ √(5/3) = x 2 -(5/3) = 0

(3x 2 −5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x 2 −5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x 4  +6x 3  −2x 2  −10x–5 = (3x 2  –5) (x 2 +2x+1)

Now, on further factorizing (x 2 +2x+1) we get,

x 2 +2x+1  = x 2 +x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by:  x= −1  and  x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

4. 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).

Dividend, p(x) = x 3 -3x 2 +x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor × Quotient + 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)

Therefore, g(x) × (x-2) = x 3 -3x 2 +3x-2

Now, for finding g(x) we will divide x 3 -3x 2 +3x-2 with (x-2)

Therefore, g(x) = (x 2 –x+1)

5. 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)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, p(x) = 3x 2 +3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x 2 +3x+3)/3 = x 2 +x+1 = q(x)

Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

Let us take an example, p(x) = x 2  + 3 is a polynomial to be divided by g(x) = x – 1.

So, x 2  + 3 = (x – 1)×(x) + (x + 3)

Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x 2  + 1 is a polynomial to be divided by g(x) = x.

So, x 2  + 1 = (x)×(x) + 1

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

## Exercise 2.4 Page: 36

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x 3 +x 2 -5x+2; -1/2, 1, -2

Given, p(x) = 2x 3 +x 2 -5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2) 3 +(1/2) 2 -5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1) 3 +(1) 2 -5(1)+2 = 0

p(-2) = 2(-2) 3 +(-2) 2 -5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x 3 +x 2 -5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax 3 +bx 2 +cx+d = 2x 3 +x 2 -5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax 3 +bx 2 +cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x 3 -4x 2 +5x-2 ; 2, 1, 1

Given, p(x) = x 3 -4x 2 +5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2 3 -4(2) 2 +5(2)-2 = 0

p(1) = 1 3 -(4×1 2 )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x 3 -4x 2 +5x-2

∴ ax 3 +bx 2 +cx+d = x 3 -4x 2 +5x-2

a = 1, b = -4, c = 5 and d = -2

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

2. 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.

Let us consider the cubic polynomial is ax 3 +bx 2 +cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x 3 -2x 2 -7x+14

3. If the zeroes of the polynomial x 3 -3x 2 +x+1 are a – b, a, a + b, find a and b.

We are given with the polynomial here,

p(x) = x 3 -3x 2 +x+1

And zeroes are given as a – b, a, a + b

∴px 3 +qx 2 +rx+s = x 3 -3x 2 +x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

Putting the values q and p.

-(-3)/1 = 3a

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b 2

-1/1 = 1-b 2

b 2 = 1+1 = 2

Hence,1-√2, 1 ,1+√2 are the zeroes of x 3 -3x 2 +x+1.

4. If two zeroes of the polynomial x 4 -6x 3 -26x 2 +138x-35 are 2 ± √ 3, find other zeroes.

Let f(x) = x 4 -6x 3 -26x 2 +138x-35

Since 2 +√ 3 and 2-√ 3 are zeroes of given polynomial f(x).

∴ [x−(2+√ 3 )] [x−(2-√ 3) ] = 0

(x−2−√ 3 )(x−2+√ 3 ) = 0

On multiplying the above equation we get,

x 2 -4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x 4 -6x 3 -26x 2 +138x-35 = (x 2 -4x+1)(x 2  –2x−35)

Now, on further factorizing (x 2 –2x−35) we get,

x 2 –(7−5)x −35  = x 2 – 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√ 3 , 2-√ 3 , −5 and 7.

Q.5: If the polynomial x 4 – 6x 3 + 16x 2 – 25x + 10 is divided by another polynomial x 2 – 2x + k, the remainder comes out to be x + a, find k and a.

Let’s divide x 4 – 6x 3 + 16x 2 – 25x + 10 by x 2 – 2x + k.

Given that the remainder of the polynomial division is x + a.

(4k – 25 + 16 – 2k)x + [10 – k(8 – k)] = x + a

(2k – 9)x + (10 – 8k + k 2 ) = x + a

Comparing the coefficients of the above equation, we get;

2k = 9 + 1 = 10

k = 10/2 = 5

10 – 8k + k 2 = a

10 – 8(5) + (5) 2 = a [since k = 5]

10 – 40 + 25 = a

Therefore, k = 5 and a = -5.

## NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

As this is one of the important topics in Maths, it comes under the unit – Algebra which has a weightage of 20 marks in the Class 10 Maths CBSE exams. The average number of questions asked from this chapter is usually 1. This chapter talks about the following,

• Introduction to Polynomials
• Geometrical Meaning of the Zeros of Polynomial
• Relationship between Zeros and Coefficients of a Polynomial
• Division Algorithm for Polynomials

Polynomials are introduced in Class 9, where we discussed polynomials in one variable and their degrees in the previous class. This is discussed in further detail in Class 10. The NCERT Solutions for Class 10 Maths for this chapter discusses the answers to various types of questions related to polynomials and their applications. We study the division algorithm for polynomials of integers, and also whether the zeroes of quadratic polynomials are related to their coefficients.

The chapter starts with the introduction of polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3

• Geometrical Meaning of the zeroes of a Polynomial – It includes 1 question having 6 different cases.
• Relationship between Zeroes and Coefficients of a Polynomial – Explore the relationship between zeroes and coefficients of a quadratic polynomial through solutions to 2 problems in Exercise 2.2, having 6 parts in each question.

Next, it discusses the following topics, which were introduced in Class 9.

• Division Algorithm for Polynomials – In this, the solutions for 5 problems in Exercise 2.3 is given, having three long questions.

## Key Features of NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

• It covers the CBSE syllabus for 2023-24 of Class 10 Maths.
• After studying these NCERT Solutions prepared by our subject experts, you will be confident of scoring well in the exams.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.

For a strong grip over the concepts, students can also make use of the other reference materials which are present at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

Disclaimer –

Dropped Topics –  2.4 Division algorithm for polynomials

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 2

Where can i get the accurate solution for ncert solutions for class 10 maths chapter 2, is it necessary to solve each problem provided in the ncert solutions for class 10 maths chapter 2, list out the concepts covered in ncert solutions for class 10 maths polynomials., leave a comment cancel reply.

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PPT slide on Polynomial (Class 10th) compiled by Rishi. ... A polynomial p(x) = ax 4 + bx3 + cx2 + dx + e of degree 4 is called a bi-quadratic polynomial. @maths ithrishi ; Value of a Polynomial at a given poin Ifp(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k inp(x), is called the value of p ...

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quadratic equation class 10th OBJECTIVES QUADRATIC EQUATION DEFINITION METHODS TO FIND THE SOLUTION OF QUADRATIC EQUATIONS DISCRIMINANT NATURE OF ROOTS EXAMAPLES @mathswithrishi QUADRATIC EQUATION A polynomial p(x) = ax2 + bx + c of degree 2 is called a quadratic polynomial, then p(x) 0 is known as a quadratic equation. e.g. 2x2 — 3x + 2 O ...

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