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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

Updated on Nov 08, 2023 06:25 PM IST | #CBSE Class 10th

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:
2x3+x2−5x+2;12,1,−2

Answer:

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

p(12)=2×(12)3+(12)2−5×12+2p(12)=14+14−52+2p(12)=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 α,β and γ

α=12β=1γ=−2

α+β+γ=12+1+(−2)=−12=−ba

Verified

αβ+βγ+γα=12×1+1×(−2)+(−2)×12=−52=ca

Verified

αβγ=12×1×−2=−1=−22=−da

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:

x3−4x2+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 α,β and γ

α=2β=1γ=1

α+β+γ=2+1+1=4=−−41=−ba

Verified

αβ+βγ+γα=2×1+1×1+1×2=5=51=ca

Verified

αβγ=2×1×1=2=−−21=−da

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 α,β and γ

α+β+γ=2αβ+βγ+γα=−7αβγ=−14

x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ=0x3−2x2−7x+14=0

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

Q3 If the zeroes of the polynomial x3−3x2+x+1 are a – b, a, a + b, find a and b.

Answer:

x3−3x2+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=±2

Therefore a = 1 and b=±2 .

Q4 If two zeroes of the polynomial x4−6x3−26x2+138x−35 x4−6x3−26x2+138x−35 are 2±3 , find other zeroes .

Answer: Given the two zeroes are

2+3 and 2−3

therefore the factors are

[x−(2+3)] and[x−( 2−3)]

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+3)]\[x−( 2−3)]=x2−(2+3)x−( 2−3)x+( 2+3)( 2−3)=x2−2x−3x−2x+3x+1=x2−4x+1

Now carrying out the polynomial division

1635919283320

Now we get x2−2x−35 is also a factor

x2−2x−35=x2−7x+5x−35=x(x−7)+5(x−7)=(x−7)(x+5)

So the zeroes are 2+3 , 2−3,7 and −5

Q5 If the polynomial x4−6x3+16x2−35x+10 is divided by another polynomial x2−2x+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.
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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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i studied 1 to 10th CBSE board abroad and 11th 12th in karnataka state board . i am a domicile of karnataka and comes under minority, income under 5 lakhs. how neet councillng will allocate my seat

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

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Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

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Domicile Certificate (proof that you are a resident of Karnataka).

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Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

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Mitali Sharma 30 January,2025
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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