NCERT Solutions for Exercise 2.4 Class 9 Maths Chapter 2

NCERT Solutions for Exercise 2.4 Class 9 Maths Chapter 2 - Polynomials

Edited By Vishal kumar | Updated on Oct 04, 2023 10:40 AM IST

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.4- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.4- Download Free PDF- NCERT Solutions for exercise 2.4 Class 9 Maths Chapter 2 Polynomials Exercise 2.4 is the part of NCERT solutions for Class 9 Maths. A polynomial expression is an equation made up of variables (or indeterminate variables), terms, exponents, and constants. If we talk about exercise 2.4 Class 9 Maths is an exercise of the chapter introduced and followed by exercise 2.3 that includes some numerical problems. Here in this exercise 2.4, we will be studying the factorisation of Polynomials.

Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

New: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9

This Story also Contains

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.4- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.4
Access Polynomials Class 9 Chapter 2 Exercise: 2.4
More About NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4
Key Features of Exercise 2.4 Class 9 Maths
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

NCERT Solutions for Exercise 2.4 Class 9 Maths Chapter 2 - Polynomials

The Class 9 Maths chapter 2 exercise 2.4 lists some basic level practice problems on the polynomials chapter that consist of factorization of higher degree Polynomials. The Class 9 Maths chapter 2 exercise 2.4 covers the topics like factorization theorem enclosed with examples. NCERT solutions for Class 9 Maths chapter 2 exercise 2.4 gives an end-to-end idea of the whole chapter.

The 9th class maths exercise 2.4 answers have been expertly crafted by subject experts, and presented in detail and easy-to-understand language. This exercise comprises a total of five questions, each with multiple parts. Students can conveniently access these class 9 maths chapter 2 exercise 2.4 solutions in PDF format, allowing offline usage without the need for an internet connection, and they are provided free of charge. Along with Class 9 Maths chapter 2 exercise 2.4, the following exercises are also present. Along with Class 9 Maths chapter 1 exercise 2.4 the following exercises are also present.

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.4

Download PDF

Access Polynomials Class 9 Chapter 2 Exercise: 2.4

Q1 (i) Determine which of the following polynomials has $(x + 1)$ a factor : $x^3 + x^2 +x + 1$

Answer:

Zero of polynomial $(x + 1)$ is -1.

If $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$

Then, $p(-1)$ must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^3+(-1)^2-1+1$

$\Rightarrow p(-1)=-1+1-1+1 = 0$

Therefore, $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$

Q1 (ii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + x^3 + x^2 +x + 1$

Answer:

Zero of polynomial $(x + 1)$ is -1.

If $(x + 1)$ is a factor of polynomial $p(x)=x^4 + x^3 + x^2 +x + 1$

Then, $p(-1)$ must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^4+(-1)^3+(-1)^2-1+1$

$\Rightarrow p(-1)=1-1+1-1+1 = 1\neq 0$

Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^4 + x^3 + x^2 +x + 1$

Q1 (iii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + 3x^3 + 3x^2 +x + 1$

Answer:

Zero of polynomial $(x + 1)$ is -1.

If $(x + 1)$ is a factor of polynomial $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$

Then, $p(-1)$ must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^4+3(-1)^3+3(-1)^2-1+1$

$\Rightarrow p(-1)=1-3+3-1+1 = 1\neq 0$

Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$

Q1 (iv) Determine which of the following polynomials has $(x + 1)$ a factor : $x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Answer:

Zero of polynomial $(x + 1)$ is -1.

If $(x + 1)$ is a factor of polynomial $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Then, $p(-1)$ must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^3-(-1)^2-(2+\sqrt2)(-1)+\sqrt2$

$\Rightarrow p(-1)=-1-1+2+\sqrt2+\sqrt2 = 2\sqrt2\neq 0$

Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Q2 (i) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = 2x^3 + x^2 - 2x - 1,\ g(x) = x + 1$

Answer:

Zero of polynomial $g(x)=x+1$ is $-1$

If $g(x)=x+1$ is factor of polynomial $p(x) = 2x^3 + x^2 - 2x - 1$

Then, $p(-1)$ must be equal to zero

Now,

$\Rightarrow p(-1)= 2(-1)^3+(-1)^2-2(-1)-1$

$\Rightarrow p(-1)= -2+1+2-1 = 0$

Therefore, $g(x)=x+1$ is factor of polynomial $p(x) = 2x^3 + x^2 - 2x - 1$

Q2 (ii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = x^3 + 3x^2 + 3x + 1, \ g(x) = x + 2$

Answer:

Zero of polynomial $g(x)=x+2$ is $-2$

If $g(x)=x+2$ is factor of polynomial $p(x) = x^3 + 3x^2 + 3x + 1$

Then, $p(-2)$ must be equal to zero

Now,

$\Rightarrow p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$

$\Rightarrow p(-2)= -8+12-6+1 = -1\neq 0$

Therefore, $g(x)=x+2$ is not a factor of polynomial $p(x) = x^3 + 3x^2 + 3x + 1$

Q2 (iii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = x^3 - 4x^2 + x + 6, \ g(x) = x - 3$

Answer:

Zero of polynomial $g(x)=x-3$ is $3$

If $g(x)=x-3$ is factor of polynomial $p(x) = x^3 - 4x^2 + x + 6$

Then, $p(3)$ must be equal to zero

Now,

$\Rightarrow p(3) = (3)^3 - 4(3)^2 + 3 + 6$

$\Rightarrow p(3) = 27-36+3+6=0$

Therefore, $g(x)=x-3$ is a factor of polynomial $p(x) = x^3 - 4x^2 + x + 6$

Q3 (i) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = x^2 + x + k$

Answer:

Zero of polynomial $x - 1$ is $1$

If $x - 1$ is factor of polynomial $p(x) = x^2 + x + k$

Then, $p(1)$ must be equal to zero

Now,

$\Rightarrow p(1) = (1)^2 + 1 + k$

$\Rightarrow p(1) =0$

$\Rightarrow 2+k = 0$

$\Rightarrow k = -2$

Therefore, value of k is $-2$

Q3 (ii) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = 2x^2 + kx + \sqrt2$

Answer:

Zero of the polynomial $x - 1$ is $1$

If $x - 1$ is factor of polynomial $p(x) = 2x^2 + kx + \sqrt2$

Then, $p(1)$ must be equal to zero

Now,

$\Rightarrow p(1) = 2(1)^2 + k(1) + \sqrt2$

$\Rightarrow p(1) =0$

$\Rightarrow 2+k +\sqrt2= 0$

$\Rightarrow k = -(2+\sqrt2)$

Therefore, value of k is $-(2+\sqrt2)$

Q3 (iii) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = kx^2-\sqrt2 x + 1$

Answer:

Zero of polynomial $x - 1$ is $1$

If $x - 1$ is factor of polynomial $p(x) = kx^2-\sqrt2 x + 1$

Then, $p(1)$ must be equal to zero

Now,

$\Rightarrow p(1) = k(1)^2 -\sqrt2(1) + 1$

$\Rightarrow p(1) =0$

$\Rightarrow k -\sqrt2 +1= 0$

$\Rightarrow k = -1+\sqrt2$

Therefore, value of k is $-1+\sqrt2$

Q3 (iv) the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = kx^2 -3 x + k$

Answer:

Zero of polynomial $x - 1$ is $1$

If $x - 1$ is factor of polynomial $p(x) = kx^2 -3 x + k$

Then, $p(1)$ must be equal to zero

Now,

$\Rightarrow p(1) = k(1)^2 -3(1) + k$

$\Rightarrow p(1) =0$

$\Rightarrow k -3+k= 0$

$\Rightarrow k = \frac{3}{2}$
Therefore, value of k is $\frac{3}{2}$

Q4 (i) Factorise : $12x^2 - 7x + 1$

Answer:

Given polynomial is $12x^2 - 7x + 1$

We need to factorise the middle term into two terms such that their product is equal to $12 \times 1 = 12$ and their sum is equal to $-7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 12x^2-3x-4x+1$ $(\because -3 \times -4 = 12 \ \ and \ \ -3+(-4) = -7)$

$\Rightarrow 3x(4x-1)-1(4x-1)$

$\Rightarrow (3x-1)(4x-1)$

Q4 (ii) Factorise : $2x^2 + 7x + 3$

Answer:

Given polynomial is $2x^2 + 7x + 3$

We need to factorise the middle term into two terms such that their product is equal to $2 \times 3 = 6$ and their sum is equal to $7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 2x^2+6x+x+3$ $(\because 6 \times 1 = 6 \ \ and \ \ 6+1 = 7)$

$\Rightarrow 2x(x+3)+1(x+3)$

$\Rightarrow (2x+1)(x+3)$

Q4 (iii) Factorise : $6x^2 +5x - 6$

Answer:

Given polynomial is $6x^2 +5x - 6$

We need to factorise the middle term into two terms such that their product is equal to $6 \times -6 =-3 6$ and their sum is equal to $5$

We can solve it as

$\Rightarrow 6x^2 +5x - 6$

$\Rightarrow 6x^2 +9x -4x - 6$ $(\because 9 \times -4 = -36 \ \ and \ \ 9+(-4) = 5)$

$\Rightarrow 3x(2x+3)-2(2x+3)$

$\Rightarrow (2x+3)(3x-2)$

Q4 (iv) Factorise : $3x^2 - x - 4$

Answer:

Given polynomial is $3x^2 - x - 4$

We need to factorise the middle term into two terms such that their product is equal to $3 \times -4 =-12$ and their sum is equal to $-1$

We can solve it as

$\Rightarrow 3x^2 - x - 4$

$\Rightarrow 3x^2 -4 x+3x - 4$ $(\because 3 \times -4 = -12 \ \ and \ \ 3+(-4) = -1)$

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

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

Q5 (i) Factorise : $x^3 - 2x^2 - x +2$

Answer:

Given polynomial is $x^3 - 2x^2 - x +2$

Now, by hit and trial method we observed that $(x+1)$ is one of the factors of the given polynomial.

By long division method, we will get

1639996302612 We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 2x^2 - x +2 = (x+1)(x^2-3x+2)+0$

$= (x+1)(x^2-2x-x+2)$

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

Therefore, on factorization of $x^3 - 2x^2 - x +2$ we will get $(x+1)(x-2)(x-1)$

Q5 (ii) Factorise : $x^3 - 3x^2 -9x -5$

Answer:

Given polynomial is $x^3 - 3x^2 -9x -5$

Now, by hit and trial method we observed that $(x+1)$ is one of the factors of the given polynomial.

By long division method, we will get

1639996323635 We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)$

$= (x+1)(x^2-5x+x-5)$

$= (x+1)(x-5)(x+1)$

Therefore, on factorization of $x^3 - 3x^2 -9x -5$ we will get $(x+1)(x-5)(x+1)$

Q5 (iii) Factorise : $x^3 + 13x^2 + 32x + 20$

Answer:

Given polynomial is $x^3 + 13x^2 + 32x + 20$

Now, by hit and trial method we observed that $(x+1)$ is one of the factore of given polynomial.

By long division method, we will get

1639996347806 We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 + 13x^2 + 32x + 20=(x+1)(x^2+12x+20)$

$= (x+1)(x^2+10x+2x+20)$

$= (x+1)(x+10)(x+2)$

Therefore, on factorization of $x^3 + 13x^2 + 32x + 20$ we will get $(x+1)(x+10)(x+2)$

Q5 (iv) Factorise : $2y^3 + y^2 - 2y - 1$

Answer:

Given polynomial is $2y^3 + y^2 - 2y - 1$

Now, by hit and trial method we observed that $(y-1)$ is one of the factors of the given polynomial.

By long division method, we will get

1639996378575 We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)$

$= (y-1)(2y^2+2y+y+1)$

$= (y-1)(2y+1)(y+1)$

Therefore, on factorization of $2y^3 + y^2 - 2y - 1$ we will get $(y-1)(2y+1)(y+1)$

Key Features of Exercise 2.4 Class 9 Maths

Comprehensive Exercise: 9th class maths exercise 2.4 answers is a comprehensive exercise that covers various topics related to polynomials.
Conceptual Clarity: The primary objective of this class 9 maths chapter 2 exercise 2.4 is to help students develop a clear understanding of polynomial factorization and related theorems.
Diverse Problem Set: exercise 2.4 class 9 maths offers a variety of problems with different levels of complexity, allowing students to enhance their skills in polynomial factorization.
Expert-Created Solutions: class 9 maths ex 2.4 Solutions to the problems are typically provided in the exercise. These solutions are crafted by subject matter experts to ensure accuracy and clarity.
PDF Availability: Students can often download a PDF version of the ex 2.4 class 9 solutions, allowing them to access and use them offline. This resource is provided at no cost.

Also Read| Polynomials Class 9 Notes

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. What is the middle term theorem?

Middle term theorem is used to find roots/zeros of equation ax² + bx + c= 0 that means we have to find that roots whose sum = -(b/a) and the product is equal to c/a

2. What is the degree of Constant Polynomial expression?

The degree of the constant polynomial is 0. Taking example is, 2 is a constant polynomial that is equal to 2x^0, so its degree is 0.

3. How many types of questions are there in chapter 2 exercise 2.4?

There are a total of 5 types of questions in the exercise 2.4

4. What is the sum of roots of polynomial 3x² +6x + 9=0?

Comparing the above polynomial with ax² + bx + c= 0, a= 3, b = 6 and c= 9, hence sum of roots = -(b/a) = -(6/3) = -2

5. What is the product of roots of the above polynomial?

Product of roots = c/a = 9/3 = 3

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Articles

High Salary Courses After Class 12 Science

Jul 26, 2024

Ophthalmology Courses After 12th - Eligibility, Duration, Institutes

Jul 26, 2024

BiPC Courses After 12th - Fees, Duration & Top Institutes

Jul 25, 2024

ICSO Sample Papers 2024-25 with Answers For Classes 1 to 10

Jul 25, 2024

NMMS Chhattisgarh Result 2024-25: Check Merit List, Cut Off Marks

Jul 25, 2024

UPTAC Counselling 2024 (Started): Document Verification, Date, Registration, Seat Allotment, Rules, Guidelines

Jul 25, 2024

Essay on Diwali for Students: Short Paragraph, 10 Lines on Deepavali

Jul 24, 2024

How to Study at Home? - Check 10 Tips to Study at Home

Jul 23, 2024

IIM Courses After 12th - Eligibility, Exams, Top Colleges

Jul 19, 2024

Upcoming School Exams

National Means Cum-Merit Scholarship

Application Date:15 July,2024 - 14 August,2024

Eligibility Criteria

Application Process

Exam Pattern

National Institute of Open Schooling 12th Examination

Late Fee Application Date:22 July,2024 - 31 July,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Late Fee Application Date:22 July,2024 - 31 July,2024

Application Process

Exam Pattern

Mock Test

Chhattisgarh Board of Secondary Education 12th Examination

Exam Date:23 July,2024 - 12 August,2024

Exam Pattern

Admit Card

Result

Chhattisgarh Board of Secondary Education 10th Examination

Exam Date:24 July,2024 - 08 August,2024

Exam Pattern

Admit Card

Result

View All School Exams

Get answers from students and experts

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is