NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.4

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 are provided here to help students to clear their concepts. These solutions are prepared by our subject expert in Math to let the students in Class 10 prepare for their board examination. These experts create NCERT Solutions for Maths so that it would help students to solve the questions based on this chapter easily. The solutions are prepared in a stepwise method so that students can learn the concepts quickly. 

The subject experts at Vedantu provide a detailed solution to each answer to the questions given in exercise 2.4 in the NCERT textbook for Class 10. The NCERT Solutions for Class 10 Maths Chapter 2 Polynomials  Exercise 2.4 are prepared by following NCERT guidelines keeping in mind that it should cover the whole syllabus accordingly. These are very helpful in scoring well on the board as well as competitive examinations.

Topics Covered

• Introduction

• Types of Polynomial

• Geometrical Meaning of Zeros of Polynomial

What is a Polynomial?

Polynomial is derived from the word poly- (meaning "many '') and -nominal meaning "term". In other words, it says many terms.

A polynomial can have constants, variables, and exponents. These constants, variables, and exponents can be combined using addition, subtraction, and multiplication but not by division. 

A polynomial may look like this:

6xy² + 2x - 5 

Here, constants are 6, -2, and 5

Variables are x and y

Exponents like 2 in y. ( y²)

Types of Polynomial

Linear Polynomial: A polynomial of degree 1 is called a linear polynomial. For example, 5x - 3.

Quadratic Polynomial: A polynomial of degree 2 is called a quadratic polynomial. For example, y² - 3.

Cubic Polynomial: A polynomial of degree 3 is called a cubic  polynomial. For example, 2- y3.

What Are Zero Polynomials?

Zeros of polynomials is a type of polynomial where all variable coefficients become zero. In other words, all the variables have a power that is equal to zero. 

Exercise - 2.4

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

1. \[\mathbf{2{{x}^{3}}+ {{x}^{2}}-5x + 2;  \dfrac{1}{2},1,-2}\]

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,

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

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

\[\text{= 0}\]

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

\[\text{= 0}\] 

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

\[\text{=}-\text{16 + 4 +10 + 2}\]

\[\text{= 0}\] 

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 \[\text{  }\!\!\gamma\!\!\text{  =}-\text{2}\].

Then,

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

\[\therefore \text{ }\!\!\alpha\!\!\text{  +  }\!\!\beta\!\!\text{  +  }\!\!\gamma\!\!\text{  =}\dfrac{-\text{b}}{\text{a}}\]  

\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\left ( \dfrac{1}{2} \right )\left ( 1 \right )+\left ( 1 \right )\left ( -2 \right )+\left ( -2 \right )\left ( \dfrac{1}{2} \right )\]

\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-5}{2}=\dfrac{c}{a}\]

\[\Rightarrow \alpha \beta \gamma =\left ( \frac{1}{2} \right )\left ( 1 \right )\left ( -2 \right )=\dfrac{-2}{2}=\dfrac{-d}{a}\]

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

2. \[\mathbf{{{x}^{3}}-4{{x}^{2}}+5x-2; 2,1,1}\]

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 \[\text{  }\!\!\gamma\!\!\text{  =1}\].

Then,

\[\Rightarrow \alpha+\beta+\gamma= 2+1+1=4=\dfrac {-(-4)}{1}=\dfrac{-b}{a}\]

\[\therefore \alpha+\beta+\gamma=\dfrac{-b}{a}\]

\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\left ( 2 \right )\left ( 1 \right )+\left ( 1 \right )\left ( 1 \right )+\left ( 1 \right )\left ( 2 \right )\]

\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =2+1+2=\dfrac{5}{1}=\dfrac{c}{a}\]

\[\therefore \alpha \beta +\beta \gamma +\gamma \alpha=\dfrac{c}{a}\]

\[\Rightarrow \alpha \beta \gamma =\left ( 2 \right )\left ( 1 \right )\left ( 1\right )=\dfrac{2}{1}=\dfrac{-d}{a}\]

\[\therefore \alpha\beta\gamma=\dfrac{-d}{a}\]

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

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 \[\alpha, \beta\:and \:\gamma\]

Then given that,

\[\Rightarrow \alpha+\beta+\gamma=\dfrac{2}{1}=\dfrac{-b}{a}\]

\[\Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{-7}{1}=\dfrac{c}{a}\]

\[\Rightarrow \alpha\beta\gamma=\dfrac {-14}{1}=\dfrac {-d}{a}\]

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 \[\text{t =1}\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{  = a}-\text{b+ a + a+b}\]

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

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

\[\Rightarrow \text{3=3a}\]

\[\therefore \text{a =1}\]

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

Now,

\[\Rightarrow \alpha \beta \gamma =\left ( 1 \right )\left ( 1-b \right )\left ( 1+b \right )\]

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

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

\[\Rightarrow \text{1}-{{\text{b}}^{\text{2}}}\text{=}-\text{1}\]

\[\Rightarrow {{\text{b}}^{\text{2}}}=2\]

\[\therefore \text{b =  }\!\!\pm\!\!\text{ }\sqrt{\text{2}}\]

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 \[{{\text{x}}^{\text{2}}}-\text{4x+1}\].

$\quad \quad\qquad \quad \,\,  x^2-2x-35$

${x^2} - 4x +1{\overline{)\,\,{x^4 -6x^3 -26x^2+138x-35}\quad\quad}}$

$\quad \quad \qquad\quad\quad \underline { {x^4} - 4x^3 + {x^2}}  $

$\quad\quad\qquad\quad \quad  0 - {2}{x^3} -27x^2+138x-35$ 

$\quad \quad \quad  \qquad  \quad \quad -2x^3 + 8x^2 -2x  $

$ \quad \quad \quad  \qquad\qquad \qquad\overline {-35x^2+140x-35} $

$\quad \quad  \quad\qquad\qquad\qquad \underline { -35x^2+140x-35 }  $

$\quad \quad \quad \quad \qquad\qquad\qquad\qquad0$

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 \[{{\text{x}}^{\text{2}}}\text{-2x+k}\]

$\quad \quad\qquad \quad \,\,  x^2-4x+(8-k)$

${x^2} - 2x +k{\overline{)\,\,{x^4 -6x^3 +16x^2-26x+10-a}\quad\quad}}$

$\quad \quad \qquad\quad\quad \underline { {x^4} - 2x^3 + k{x^2}}  $

$\quad\quad\qquad\quad \quad\,\,  0 - {4}{x^3} + (16-k)x^2-26x$ 

$\quad \quad \quad  \qquad  \quad \quad \,\,-4x^3 +\text{           } 8x^2 -4kx  $

$ \quad \quad \quad  \qquad\qquad \qquad\qquad\overline {(8-k)x^2-(26-4k)x+10-a} $

$\quad \quad  \quad\qquad\qquad\qquad \qquad\underline { (8-k)x^2- (16-2k)x+ (8k-k^2) }  $

$\quad \quad \quad \quad \qquad\qquad\qquad\qquad \underline  (-10+2k)x+(10-a-8k+k^2)$

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 Exercise 2.4

Polynomials are nothing but a mathematical expression that comprises of coefficients or variables. Polynomials always consist of positive integer exponents. Sounds complicated? Don’t worry. Download Vedantu’s PDF on NCERT Chapter 2 Class 10 Maths on Polynomials Exercise 2.4 and get all your doubts cleared seamlessly. The solutions that are provided to you are very easily understandable. Try it yourself to improve your studying process. 

Clear the Concept First

Clearing the concepts of Chapter 2 Maths Class 10 that deals with Polynomials is essential for any student. This is because, without a proper understanding of what polynomials are, it is impossible to solve these problems.  It is essential to learn how to resolve a complicated set of numbers into a simple polynomial for any problem-solving. Vedantu’s notes on Ch 2 Maths Class 10 include all questions and their smart solutions that help any student learn how to solve the problems. Download the PDF On Polynomials Class 10 now.

Practice Problems

Vedantu’s notes on the Polynomial Class 10 Chapter comprise of all the probable questions that can be a part of your question paper in any examination. The solutions are also provided with several tricks to help students get a clear idea about the concepts. Now, no one can stop you from being excellent in Maths. 

Knowing About Polynomials Is Inevitable

If a student doesn’t get a precise idea of what polynomials are in 10th standard, it is sure to get very difficult for them to solve difficult math questions in future classes. Vedantu’s notes will effectively help you to prepare yourself for the next term. The Chapter that has been curated by the CBSE board is sure to give you all the knowledge about Polynomials and Vedantu will help you get a grip on the concept with very little effort. 

No Need to Put In Too Much Effort

Vedantu’s Notes on Chapter 2 Maths Class 10 makes the process of grasping the concept of polynomials very simple. A thorough practice of the solutions just once or twice will make all the difference. You can be sure of the fact that noting will be asked in the question paper outside the concepts and questions solved in the notes provided by Vedantu. The notes are for you and Vedantu knows what you need.

Facts

• An expression of the form p(x) = a0 + a1x + a2x2 + a3x3 + ……. + anxn is called a polynomial, where a0, a1, a2, a3,……., an are real numbers, called coefficients of the polynomial, and n is non-negative integer.

Example: x + 2, 3x2 + √5x - √6 etc.

• The highest power of the variable appearing in the polynomial is called its degree. For example,

1. degree of x2 + 2x + 3 is 2.

2. degree of √3x + 5 is 1

• The degree of the zero polynomial is not established.

• A linear polynomial is one degree of the polynomial. For example: x +1, 2x + 5, etc. Its general form is ax + b, where a, b and c are real numbers and a≠ 0.

• A polynomial that is of a degree two is called a quadratic polynomial. 

For example: x2 + x + 2, 2x2 - 3x + 4, etc. Its general form is ax2 + bx + c, where a, b and c are real numbers and a ≠ 0.

•  A polynomial that is of a degree 3 is called a cubic polynomial. 

For example: x3 + x2 + x + 1, 2x3 + x2 – 3x – 4, etc. Its general form is ax3 + bx2 + cx + d, where a, b, c and d are real numbers and a ≠ 0.

•  A biquadratic polynomial is a polynomial of degree 4. 

For example: x4 + x3 + 5x2 + 6x + 2, etc. Its general form is ax4 + bx3 + cx2 + dx + e, where a, b, c, d and e are real numbers and a ≠ 0.

• A real number is said to be a zero of a polynomial.  p(x), if p(k) = 0, where k is the real number.

For example: if p(x) = x2 – 4x + 4, and p(2) = (2)2 – 4(2) + 4 = 4 – 8 + 4 = 0, then 2 is a zero of polynomial p(x).

• Remainder Theorem: To understand the remainder theorem, we must have knowledge about factors and multiples,  long division algorithms. 

If f(x) is any polynomial of a degree 1 and f(x) is divided by the linear polynomial x – a, then the remainder is f(a).

• Factor Theorem: Sometimes a polynomial having values that are unknown and one of its factors are provided and we have to calculate the value of that unknown value. Sometimes, a linear polynomial is given and we have to verify whether it is a factor of given polynomial f(x)of degree greater than 1 or not. For solving such types of problems factor theorem is required. 

In this segment, we will study division algorithms for polynomials.

If any two polynomials like p(x) and g(x) along with one polynomial of the two is not equal to 0 i.e.,  g(x) ≠ 0, then we can ascertain the polynomials r(x) and q(x) such that p(x) = g(x) * q(x) + r(x) = 0 or degree of r(x) < degree of g(x). The outcome of the above numerical is known as the Division Algorithm for polynomials. 

In other words, we can say that when p(x) divided by g(x), we get q(x) as quotient and r(x) as remainder. 

Thus, Dividend = Divisor x Quotient + Remainder

Note: If r(x) = 0, then g(x) is a factor of f(x).

Method of Division Polynomials

1. Organize the terms of the dividend and the divisor in descending order of their degrees.

2. The first term of the dividend must be divided by the first term of the divisor to obtain the first term of the quotient.

3. Multiply the whole divisor by the first term if the quotient and subtract the result obtained from the dividend.

4. Observe the remainder as the new dividend and proceed as earlier.

5. Repeat the procedure till a remainder is derived which is either 0 or whose degree is less than that of the divisor. 

Applications of Division Algorithm

1. Verifying the division algorithm for polynomials

Let us consider the following example.

Let us divide  the polynomial f(x) = x3 - 3x2 + 2x - 5 by the polynomial g(x) = x2 -2x - 3 and verify the 

division algorithm.

    __________________________

x2 -2x - 3 )  x3 - 3x2 + 2x - 5     ( x-1

            x3 - 2x2 - 3x 

      ________________________

                  - x2 + 5x - 5

                  -  x2 + 2x +3

    __________________________

                              3x - 8

∴ Quotient: q(x) = x-1 and remainder r(x)  = 3x -8.

     Now, Quotient x Divisor + Remainder

          =  (x-1) (x2 - 2x - 3) + 3x - 8

          =  x3 -2x2 - 3x - x2 + 2x + 3 + 3x - 8

          =  x3 - 3x2 + 2x - 5 = Dividend

∴The division algorithm is verified.

2. To find the Quotient and Remainder using a division algorithm. Dividing f(x) = x3 - 6x2 + 11x - 5 and g(x) = x+ 1

Sol. Here, degree of f(x) = 3 & degree of g(x) = 1

∴ Degree of quotient g(x) = 3 -1 = 2

  & degree of remainder r(x) = 0

Let q(x) = ax2 + bx + c & r(x) = k 

By division algorithm, we have

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

=>         x3 - 6x2 + 11x - 5 = (ax2 + bx + c )(x + 1)  + k

=> x3 - 6x2 + 11x - 5 = ax3 + (a + b) x2 + (b + c)x + (c + k)

Equating the quotients of various powers of x on both sides, we get

1 = a (equating coefficients of x3)

-6 = a + b => b = -7 (equating coefficient of x2)

11 = b + c => c = 18 (equating coefficient of x1)

& -5 = c + k => k = -23 (equating constant terms)

∴ q(x) = ax2 + bx + c = x2 - 7x + 18

      & r(x) = -23.

3. To check whether a given polynomial is a factor of the other polynomial by applying a division algorithm. Checking whether g(x) = x2 - 2x + 1 is a factor of the polynomial f(x) = x4 -4x3 + 6x2 - 4x + 1.

Sol. Here degree of f(x) = 4 & degree of g(x) = 2

∴degree of quotient q(x) = 4 -2  & remainder r(x) will have degree 1 or less .

Let q(x) = ax + bx + c & r(x) = mx + n 

using division algorithm , we have f(x) = q(x)g(x) + r(x)

=> x4 - 4x3 + 6x2 - 4x + 1 = (ax2 + bx + c) (x2 -2x + 1) _ mx + n

=> x4 - 4x3 + 6x2 - 4x + 1 = ax4 + ( -2a + b) x3 + (a -2b + c) x2 + (b - 2c + m) x + (c + n)

Equating the coefficients of x4, x3, x2, x1 & constant terms on both sides of the equation 

respectively, we get

x4 → 1 = a

x3 → -4 = -2a + b = -2 1+ b => b = -2

x2 → 6 = a - 2b + c =1 - 2 (-2) + c  => c = 1

x → -4 = b - 2c + m = -2 -2 1 + m => m = 0

constant → 1 = c + n = 1 + n => n = 0

∴Quotient, q(x) = ax2 + bx + c = x2 -2x + 1 & remainder, r(x) = mx + n = 0x + 0  = 0

As, r = 0, g(x) is a factor of f(x).

4. To find the remaining zeroes of a polynomial when some of its zeroes are given. Consider the polynomial f(x) = 2x4 - 3x3 - 3x2 - 6x -2, if two of its zeroes are √2 & - √2, find the other zero.

Sol. If x = is a zero of a polynomial f(x), then x - is a factor of f(x)

        As, √2 & - √2 are zeroes of f(x)

        ∴(x - √2) & ( x + √2) are factors of f(x)

        ∴(x - √2)  ( x + √2), i.e x2 - 2 is a factor  of f(x)

        To find the other zeroes of f(x), we divide f(x) by g(x) = x2 - 2.

          ____________________________

x2 - 2 ) 2x4 - 3x3 - 3x2 - 6x -2    ( 2x2 - 3x + 1

        2x4- 4x2

_________________________________

    - 3x3 + x2 - 6x -2 

  - 3x3           + 6x

________________________

        x2           -2

                 x2          -2

___________________________

                0

By division algorithm, we have

2x4 - 3x3 - 3x2 - 6x -2  = (x2 - 2) (2x2 - 3x + 1)

= (x2 - 2) (2x2 - 2x - x + 1)

= (x2 - 2) {2x(x-1) - (x - 1)}

= (x - √2) (x + √2) (x - 1) (2x -1)

Hence, the zeroes of the given polynomial apart from √2 & - √2 are 1 and ½ .

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NCERT Solutions Class 10 Maths All Chapters

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solutions Class 10 Maths Chapter 2 Exercises