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

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

Edited By Ramraj Saini | Updated on Nov 08, 2023 06:25 PM IST | #CBSE Class 10th
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NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 Polynomials

NCERT Solutions for class 10 maths ex 2.4 Polynomials is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This class 10 ex 2.4 deals with polynomial and its types, algebraic expressions, degree of a polynomial expression, graphical representation of the polynomial equations, factorization, relationship between zeroes and coefficient of a polynomial.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 Polynomials
  2. Download PDF of NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 Polynomials
  3. Access Exercise 2.4 Class 10 Maths Answers
  4. More About NCERT Solutions for Class 10 Maths Exercise 2.4:
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 2.4
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 2.4 Class 10 Maths Chapter 2 - Polynomials
NCERT Solutions for Exercise 2.4 Class 10 Maths Chapter 2 - Polynomials

Mathematically, the number of terms and the degree of the polynomial, polynomial can be classified into many types. Based on the number of terms, polynomials are classified as Monomial, Binomial and Trinomial. Also based on Degree, polynomials are classified as Linear Polynomial, Quadratic polynomial and Cubic Polynomial. Along with the Class 10 Maths chapter, 2 exercise 2.4 the following exercises are also present. Students should practice all problems discussed in these exercises to command the concepts.

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Download PDF of NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 Polynomials

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Access Exercise 2.4 Class 10 Maths Answers

Polynomials Class 10 Maths Chapter 2 Excercise: 2.4

Q1 (1) Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
2x^3 + x^2 - 5x +2 ; \frac{1}{2} , 1 , -2

Answer:

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

\\p(\frac{1}{2})=2\times \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{2}-5\times \frac{1}{2}+2\\ p(\frac{1}{2})=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\\ p(\frac{1}{2})=0

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

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

p(1) = 0

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

p(-2) = -16 + 4 + 10 + 2

p(-2) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax 3 + bx 2 + cx + d, we have

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

The roots are \alpha ,\beta \ and\ \gamma

\\\alpha=\frac{1}{2}\\ \beta =1\\ \gamma =-2

\\\alpha+\beta +\gamma \\ =\frac{1}{2}+1+(-2)\\ =-\frac{1}{2}\\ =-\frac{b}{a}

Verified

\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}\\ =\frac{-5}{2}\\ =\frac{c}{a}

Verified

\\\alpha\beta\gamma \\=\frac{1}{2}\times 1\times -2\\ =-1 \\=-\frac{2}{2}\\ =-\frac{d}{a}

Verified

Q1 (2) Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

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

Answer:

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

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

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

p(-2) = 0

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

p(1) = 1 - 4 + 5 - 2

p(1) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax 3 + bx 2 + cx + d, we have

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

The roots are \alpha ,\beta \ and\ \gamma

\\\alpha=2\\ \beta =1\\ \gamma =1

\\\alpha+\beta +\gamma \\ =2+1+1\\ =4\\ =-\frac{-4}{1}\\=-\frac{b}{a}

Verified

\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =2\times 1+1\times 1+1\times 2\\ =5\\ =\frac{5}{1}\\ =\frac{c}{a}

Verified

\\\alpha\beta\gamma \\=2\times 1\times 1\\=2 \\=-\frac{-2}{1}\\=-\frac{d}{a}

Verified

Q2 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

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

\\\alpha +\beta +\gamma =2\\ \alpha \beta +\beta \gamma +\gamma \alpha =-7\\ \alpha \beta \gamma =-14

\\x^{3}-(\alpha +\beta +\gamma )x^{2}+(\alpha \beta +\beta \gamma +\gamma \alpha)x-\alpha \beta \gamma =0\\ x^{3}-2x^{2}-7x+14=0

Hence the required cubic polynomial is x 3 - 2x 2 - 7x + 14 = 0

Q3 If the zeroes of the polynomial x^3 - 3 x^2+ x +1 are a – b, a, a + b, find a and b.

Answer:

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

The roots of the above polynomial are a, a - b and a + b

Sum of the roots of the given plynomial = 3

a + (a - b) + (a + b) = 3

3a = 3

a = 1

The roots are therefore 1, 1 - b and 1 + b

Product of the roots of the given polynomial = -1

1 x (1 - b) x (1 + b) = - 1

1 - b 2 = -1

b 2 - 2 = 0

b=\pm \sqrt{2}

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

Q4 If two zeroes of the polynomial x^4 - 6x ^3 - 26 x^2 + 138 x - 35 x^4 - 6x ^3 - 26 x^2 + 138 x - 35 are 2 \pm \sqrt 3 , find other zeroes .

Answer: Given the two zeroes are

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

therefore the factors are

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

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

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

Now carrying out the polynomial division

1635919283320

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

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

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

Q5 If the polynomial x ^4 - 6x ^3 + 16 x^2 - 35 x + 10 is divided by another polynomial x^2 - 2x + k , the remainder comes out to be x + a, find k and a.

Answer: The polynomial division is carried out as follows

1635919302615

Given the remainder =x+a

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

now equating the coefficient of x

2k-9=1

which gives the value of k=5

now equating the constants

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

Therefore k=5 and a=-5

More About NCERT Solutions for Class 10 Maths Exercise 2.4:

Exercise 2.4 Class 10 Maths – consists of five problems in this section each containing a few sub-questions. In exercise 2.4 Class 10 Maths question 1 has two subsections where we need to check and demonstrate the relationship between the zeroes and the coefficients in each case. The remaining questions also support the zeros of a polynomial. And the last sum is based on the concept of division of a polynomial. The NCERT solutions for Class 10 Maths exercise 2.4 mainly focused on the zeroes of the polynomial, the relationship between the zeroes, the coefficients of the polynomial and the Division Algorithm. Two questions related to the division algorithm are given in exercise 2.4 Class 10 Maths.

Benefits of NCERT Solutions for Class 10 Maths Exercise 2.4

  • NCERT solutions for Class 10 Maths exercise 2.4 helps in solving and revising all questions of these exercises.
  • If you go through the NCERT solution for Class 10th Maths chapter 2 exercise 2.4, you will be able to get more marks if you practice it thoroughly.
  • Exercise 2.4 Class 10 Maths, is based on zeroes of the polynomial, the relationship between the zeroes, the coefficients of the polynomial and the division Algorithm which are important concepts of the chapter. also find Polynomials class 10 Notes here and use these notes to revise all the concepts discussed in this chapter.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What are terms, degrees and exponents in a polynomial 6x^2 - 5x - 18 ?

This ex 2.4 class 10 discussed concepts of  terms, degrees and exponents in a polynomial. Practice the problems discussed in it to command the concepts. as per the above problem:

Terms : 6x^2, 5x, 18 

Degree : 2 

Exponents : 2 and 1 

2. What is a quadratic polynomial?

The class 10 maths ex 2.4  explains the concepts related to quadratic polynomial. A quadratic polynomial is nothing but a polynomial with degree two. The general form of a quadratic polynomial is ax^2 + bx + c = 0 where a, b, c are real numbers 

3. Check whether the polynomial 5x^2 + 6x + 54 = 0 a quadratic polynomial.

The concepts related to quadratic polynomial are discussed in class 10 ex 2.4. Practice the problems discussed in this exercise to command these concepts. as per this problem the maximum index here is 2 . Also it is in the form of ax^2 + bx + c = 0. 

Therefore 5x^2 + 6x + 54 = 0 is a quadratic polynomial. 

4. Factorize the polynomial n^2 - 10n + 24 .

n^2 - 10n + 24 = n^2 - 4n - 6n + 24 

= n(n - 4) - 6(n - 4) 

= (n - 4)(n - 6) 

5. Is 9 a polynomial?

The number 9 can be written as 9x^0 which is a polynomial. Therefore 9 is a polynomial.

6. What is the division algorithm according to NCERT solutions for Class 10 Maths chapter 2 exercise 2.4 ?

If p(x) and g(x) are any two polynomials where g(x) ≠ 0, p(x) = g(x) × q(x) + r(x).

That is Dividend = Divisor × Quotient + Remainder

7. How many questions and what sort of questions are covered within the NCERT solutions for Class 10 Maths chapter 2 exercise 2.4?

NCERT solutions for Class 10 Maths chapter 2 exercise 2.4  consists of five problems and the questions are based on the concept of the division algorithm, to verify and prove the relationship between the zeroes, the coefficients and the zeros of a polynomial. 

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