NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.4) Exercise 2.4

Mathematics is a subject that all students are required to study during their school years. Strong mathematical abilities can help students in a variety of ways throughout their lives. Additionally, it is crucial in assisting students in understanding the ideas covered in other subjects like science. Therefore, in addition to this, having a strong understanding of mathematics can help students succeed in other fields as well. The concepts learned in Class 8 are continued in Mathematics for Class 9. In Class 9, students are introduced to a number of new concepts, and many of these concepts are applied in Class 10 as well as in subsequent classes. Therefore, a thorough understanding of mathematics in Class 10 will not only help students achieve high test scores but also benefit them in subsequent classes. Many students find it interesting to study the concepts taught in mathematics in Class 9. They also carefully study the topics. Achieving high marks in the yearly exams, though, requires more than just reading. Numbers are central to the subject of mathematics. For this reason, getting good grades in mathematics calls for a lot of constant practice. In order to prepare for exams, experts advise students to practice NCERT solutions such as the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. Extramarks offers the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 with this in mind so that students can practice a lot and perform better on the annual examinations.

One of the most well-known educational boards in India is the Central Board of Secondary Education (CBSE). The board oversees the administration of the board exams in Class 10, which is considered a crucial part of a student’s academic career. For students, Class 9 is a very important year because it lays the groundwork for the challenging concepts they will learn in Class 10. Class 9 students have to put in a lot of work to get ready for the annual exam. For a subject like Mathematics, students need to practice answering questions a lot in order to do well on the test. According to CBSE mandates, students are required to use the textbooks published by the National Council of Educational Research and Training (NCERT). The NCERT publishes the Class 9 Mathematics textbooks. The Class 9 NCERT mathematics textbook contains 15 chapters. Students preparing for the yearly exams should read the NCERT books fully. After finishing each chapter, students should attempt to solve the questions in the book, including those found in the examples and exercises. In order to check their answers, students need to know the correct answers to the questions. Therefore, Extramarks has provided the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. Even if students are attempting to solve the problems for the first time, Extramarks’ step-by-step NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 will help them fully understand the problems. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 by Extramarks also includes alternate, simpler solutions to some problems to make it easier for students to learn.

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 (Ex 2.4)

On the Extramarks website and mobile app, students can access the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 in PDF format. When students have thoroughly studied and revised all of the topics in the chapter, they should immediately begin to practice the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4.

Class 9 Maths Chapter 2 Exercise 2.4

Class 9 Maths Chapter 2 Exercise 2.4 deals with the topic of Factorisation of Polynomials. This exercise includes five questions in total. These questions are not only important for the exams but also for understanding subsequent concepts. Students should try to completely answer all these questions to be thorough with the concepts. For help in solving the questions, they can refer to the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4.

Class 9 Maths Chapter 2 Exercise 2.4

The problems in Exercise 2.4 Class 9 Math can be very challenging for students to solve. Students need to learn the right approach to solving these problems. The exercises from the NCERT books contain questions with answers. However, just giving students the answers is insufficient because students also need to learn how the answers were arrived at. Students can learn how to respond to any question posed to them if the methods and procedures are made clear. The Extramarks-provided NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4, is an example of the kind of study tool that enables students to understand and learn about the solutions and methods from scratch.

Polynomial

Polynomials are one of the most important concepts in Mathematics. Polynomial is a type of algebraic expression that has a wide range of applications. It is important for students to thoroughly study Chapter 2 to understand the different types of concepts associated with polynomials, such as the Remainder Theorem, Factorisation, etc.

Factor Theorem:

The factor theorem is a theorem that helps in determining if an algebraic expression is a factor of a given polynomial. It is applied when factoring a polynomial. It is also used to find the roots of a polynomial. It is essential that students solve Class 9 Maths Ex 2.4 to thoroughly understand the factor theorem. They can also use the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4, for assistance in solving the problems.

NCERT Solutions for Class 9 Maths Chapter 2 Exercises

Chapter 2 of the NCERT book for Class 9 includes five exercises in total. Students attempting to solve these exercises often need reliable NCERT solutions for help. Extramarks provides comprehensive and reliable NCERT solutions, such as the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. Even though many platforms provide these solutions, some experts have doubts about their legitimacy. Students must use reliable study materials because scoring well in exams is crucial. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4, and other exercises provided by Extramarks are explained in simple language. With the help of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4, students can learn the proper format for structuring their answers. They offer the most effective and simple solutions to mathematical problems.

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NCERT Solutions for Class 9

Understanding the various concepts taught in Class 9 Mathematics requires receiving the appropriate instruction. The pressure of taking exams only makes it harder to succeed in a subject like Mathematics. At Extramarks, students can get NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 from top subject experts in addition to using other types of help to fully comprehend the concepts in a particular topic. Students can access the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 from reputed subject experts by using the carefully created learning resources that Extramarks provides. Students can gain confidence in their study strategy and for the upcoming exams with the help of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. To help students advance their education, Class 9 includes a variety of assessments and tests at the school level. To succeed in their exams, however, students require the assistance of qualified experts. Exam preparation might be aided by having access to NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 from top subject experts.

CBSE Study Materials for Class 9

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CBSE Study Materials

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Q.1

Determine which of the following polynomials has x+1 afactor:i x3+x2+x+1ii x4+x3+x2+x+1iii x4+3x3+3x2+x+1iv x3−x2−2+2x+2

Ans
(i) Let P(x) = x3 + x2 + x + 1 is divided by x+1 i.e., x = –1 if x + 1 = 0. So, P(–1) = (–1)3 + (–1)2 + (–1) + 1 = –1 + 1 –1 + 1 = 0 Since, remainder is 0. So, (x + 1) is a factor of the given polynomial. (ii) Let P(x) = x4 + x3 + x2 + x + 1 is divided by x+1
i.e., x = –1 if x + 1 = 0. So, P(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1 = 1 – 1 + 1 – 1 + 1 = 1 Since, remainder is 1. So, (x + 1) is not a factor of the given polynomial.
(iii) Let P(x) = x4 + 3x3 + 3x2 + x + 1 is divided by x+1
i.e., x = –1 if x + 1 = 0. So, P(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 Since, remainder is 1. So, (x + 1) is not a factor of the given polynomial.

(iv) Let P(x)=  x3−x2−(2+2)x+2 is divided by x+1 i.e., x =–1 if x + 1 = 0.So, P(–1) =(–1)3−(–1)2−(2+2)(–1)+2   =−1 −1 +2 +2 + 2   = 22≠0Since, remainder is 22. So, (x + 1) is not a factor of the given polynomial.

Q.2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3

Ans

( i ) Let P( x )=  2 x 3 + x 2 – 2x–1 is divided by g( x )=x+1 i.e.,x=–1. So, P( –1 ) =2 ( –1 ) 3 + ( –1 ) 2 – 2( –1 )–1   =−2 +1 +2 −1=0 Since, remainder is 0. So, g( x ) is a factor of given polynomial  P( x ). ( ii ) Let P( x )=   x 3 + 3 x 2 + 3x+1 is divided by g( x )=x+2 i.e.,x=–2. So, P( –2 ) = ( –2 ) 3 + 3 ( –2 ) 2 + 3( –2 )+1   =− 8+12−6+1=−1 Since, remainder is −1. So, g( x ) is not a factor of given polynomial  P( x ). ( iii ) Let P( x )=   x 3 –4 x 2 +x+6 is divided by g( x )=x−3 i.e.,x=3. So, P( 3 )= ( 3 ) 3 –4 ( 3 ) 2 +( 3 )+6          =27−36+3+6=0 Since, remainder is 0. So, g( x ) is a factor of given polynomial  P( x ). 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Q.3

Find the value of k, if x-1 is a factor of px in each of the following cases:i px = x2+x+kii px = 2x2+kx+2iii px = kx2−2x+1iv px = kx2−3x+k

Ans

 (i) p(x)=x2+x+kSince, x−1 is a factor of p(x)=x2+x+k.So, by remainder theorem,    p(1)=012+1+k=0⇒k=−2Thus, the value of k is −2.ii px=2x2+kx+2Since, x−1 is a factor of px=2x2+kx+2.So, by remainder theorem,                        p(1)=02(1)2+k(1)+2=0⇒k=− 2−2Thus, the value of k is −(2+2).iii px=kx2−2x+1Since, x−1 is a factor of px=kx2−2x+1.So, by remainder theorem,                     p(1)=0k(1)2−2(1)+1=0⇒k=−1+ 2Thus, the value of k is (2−1).  

iv px=kx2−3x+kSince, x−1 is a factor of px=kx2−3x+k.So, by remainder theorem,                 p(1)=0k(1)2−3(1)+k=0⇒     2k=3⇒       k=32Thus, the value of k is 32.

Q.4 Factorise :
(i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4

Ans
(i) 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1
= 4x (3x–1) – 1(3x–1)
= (3x–1)(4x–1) (ii) 2x2 + 7x + 3= 2x2 + 6x + x + 3
= 2x (x+3) + 1(x+3)
= (x+ 3) (2x+1) (iii) 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x (2x+3) – 2(2x + 3)
= (2x + 3)(3x– 2) (iv) 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x (3x – 4) + 1(3x – 4)
= (3x – 4) (x+1)

Q.5 Factorise:
(i) x3 – 2x2 – x + 2 (ii) x3 – 3x2 – 9x – 5
(iii) x3+13x2+32x +20 (iv) 2y3 + y2 – 2y – 1

Ans
(i) Let p(x) = x3− 2x2− x + 2
All the factors of 2 have to be considered.
These are ± 1, ± 2.
By trial method, we get
p(2) = (2)3 − 2(2)2− 2 + 2
= 8 – 8 – 2 + 2
= 0
Therefore, (x − 2) is a factor of polynomial p(x).
Let us find the quotient on dividing x3− 2x2− x + 2 by x − 2.

x3− 2x2−x+2 is divided by x−2                         x2−1              x−2   x3− 2x2−x+2                      ±x3∓2x2       ¯                                   −x+2                                    ∓x±2¯                                         0       ¯          Since,                  Dividend=Divisor×Quotient+Remainderx3− 2x2−x+2=(x−2)(x2−1)+0  =(x−2)(x+1)(x−1)

(ii) Let p(x) = x3 – 3x2 – 9x – 5
All the factors of 5 have to be considered.
These are ± 1, ± 5.
By trial method, p(5) = (5)3 – 3(5)2 – 9(5) – 5
= 125 – 75 – 45 – 5
= 0
Therefore, (x − 5) is a factor of polynomial p(x).
Let us find the quotient on dividing x3 – 3x2 – 9x – 5 by x − 5.

x3– 3x2– 9x– 5 is divided by x−5         x2+2x+1x−5   x3– 3x2– 9x– 5        ±x3∓5x2  ¯                      2x2– 9x –5                  ±2x2∓10x  ¯                  x−5                   ±x∓5  ¯                            0    ¯          Since,                        Dividend=Divisor×Quotient+Remainderx3– 3x2– 9x– 5=(x−5)(x2+2x+1)+0         =(x−5)(x+1)(x+1)

(iii) Let p(x) = x3+ 13x2 + 32x + 20
All the factors of 20 have to be considered.
These are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20.
By trial method, p(−1) = (−1)3 + 13(−1)2 +32(−1)+ 20
= −1+13−32 + 20
= 0
Therefore, (x + 1) is a factor of polynomial p(x).
Let us find the quotient on dividing x3+ 13x2 + 32x + 20 by x + 1.

x3+ 13x2+ 32x+ 20 is divided by x+1                         x2+12x+20              x+1   x3+ 13x2+ 32x+ 20                      ±x3±x2 ¯                                  12x2+32x+20                               ±12x2±12x¯                                                 20x+20                            20x+20¯                             0    ¯Since,                                 Dividend=Divisor×Quotient+Remainderx3+ 13x2+ 32x+ 20=(x+1)(x2+12x+20)+0       =(x+1)(x+2)(x+10)

(iv) Let p(y) = 2y3 + y2 – 2y – 1
All the factors of 2 have to be considered.
These are ± 1, ± 2.
By trial method, p(1) = 2(1)3 + (1)2 − 2(1) – 1
= 2 + 1 – 2 – 1
= 0
Therefore, (y − 1) is a factor of polynomial p(y).
Let us find the quotient on dividing 2y3 + y2 – 2y – 1 by y − 1

2y3+y2– 2y– 1 is divided by y−1                   2y2+y+1           y−1   2y3+y2– 2y– 1                  ±2y3∓2y2¯                               y2−2y–1                             ±y2∓y¯                                      y−1   ±y∓1   ¯                                      0        ¯Since,                  Dividend=Divisor×Quotient+Remainder2y3+y2– 2y–1=(y−1)(2y2+y+1)+0  =(y−1)(y+1)(2y+1)

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1. Are there NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 available on Extramarks' website?

Yes, both on its website and in its mobile app, Extramarks has the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. To perform better on the test, students should practice NCERT exercises regularly.

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Understanding the question types beforehand and identifying various important questions can both be accomplished by practising the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. More practice with NCERT exercises can help in answering questions faster. They will ultimately be able to finish the entire Mathematics exam question paper in the allotted time due to this.

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All five questions in Exercise 2.4 are answered in the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4. To achieve the highest possible score on the exam, students can find all the steps required to solve the questions through the Class 9 Maths Chapter 2 Exercise 2.4 Solution provided by Extramarks.

4. How can the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4 help in achieving high marks in the annual exams?

When preparing for the Class 9 annual exams, NCERT solutions are the most beneficial. The majority of exam questions are based on the NCERT textbook, which is recommended by the CBSE. To answer these questions correctly and receive full marks, students should refer to the step-by-step solutions in the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4.