NCERT Solutions for Class 9 Mathematics Chapter 2 – Polynomials

Mathematics is based on application  and its application in the right manner  strengthens the conceptual understanding of the topic. . The Polynomial chapter is considered one of the important chapters in Algebra. Hence, it plays a crucial role in Class 9 Mathematics.  This chapter covers topics like polynomials in one variable, remainder theorem, algebraic identities which carries  great weightage in Mathematics syllabus. One can find all the concepts, theories  and NCERT Solutions along with a lot of solved examples on our Extramarks official website. 

NCERT Solutions for Class 9 Mathematics Chapter 2 can be availed from the Extramarks website by all the students studying in Class 9.  You can find a detailed explanation and complete solution to all the questions in the NCERT textbook in our NCERT Solution. Designed as per  the  latest CBSE syllabus 2021-2022, we provide chapter explanations which are easy to understand for the students. Therefore, students can    take full advantage of the NCERT Solutions for Class 9 Mathematics Chapter 2 present on the Extramarks website for their school exam preparation as well as other competitive exams.

Apart from NCERT solutions, Extramarks website has an abundance of  study materials  like CBSE short notes, past year question papers, NCERT books and  much more. Find NCERT Solutions for Class 9 Mathematics Chapter 2 on Extramarks website and gain mastery on all the core concepts covered in Chapter 2 Polynomials.

Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 2 

In order to have a better understanding of  Class 9 Mathematics Chapter 2, one must understand  Chapter 1 Number systems. The core concepts covered in earlier  chapters will be used in Chapter 2. So, it is necessary that students know properly how to identify different sets of numbers and their rationalisation. The topics which will be covered in Chapter 2 will form a basis of  algebra.  . Hence, polynomials are considered the backbone of algebra and will go a long way in the curriculum.

Extramarks website provides NCERT Solutions for Class 9 Mathematics Chapter 2. It begins with a basic introduction of Whole Numbers, Integers, Rational Numbers which we have already covered in the previous chapters  followed by discussion of linear, quadratic and cubic polynomials. You will also find detailed applications of the operations on real numbers. Students who study  in a  systematic manner will  not only build a strong  Algebra foundation but great mathematical abilities  to write various Competitive examinations.

The following key topics are covered in NCERT Solutions for Class 9 Mathematics Chapter 2:

Exercise Topic
2.1 Introduction
2.2 Polynomials in one variable
2.3 Zeros of a Polynomial
2.4 Remainder Theorem
2.5 Factorisation of Polynomials
2.6 Algebraic identities 

This chapter requires proper comprehension of the Algebraic expressions as well as factorisation and needs students to recall the wide range of identities they have read so far.

2.1 Introduction

Students have learned about the addition, subtraction, multiplication and division of algebraic expressions in their earlier classes. This chapter entirely deals with algebraic expressions called polynomials and all the concepts related to it. Mathematics Class 9 Chapter 2 also briefly deals with algebraic identities and their application in evaluating some given expressions. 

2.2 Polynomials in one variable

This section starts with the recalling of a variable and how it is denoted In algebraic expressions. You would also get to know how to write an expression for a constant.

The various examples provided in our NCERT Solutions for Class 9 Mathematics Chapter 2 helps one in understanding the difference between the constant and the variable. You also get to know about the steps to follow to recognize a polynomial.

Students can distinguish between a letter of a constant and a letter denoting a variable. You will also get to know that each term of a polynomial is represented by a coefficient and a constant polynomial is called a zero polynomial.   

2.3 Zeros of a Polynomial

In this section, you will actually learn how to find the zeros of a polynomial using the examples provided in the  chapter. You will get to know the various conventions of real numbers and a list of observations related to polynomials.

After completing zeros of polynomials, you will know that every linear polynomial has one and only one zero but a polynomial can have more than one zero. In short it gives a clear cut idea on solving polynomial equations.

2.4 Remainder Theorem

Recalling the basics of division, when we divide a number by a non-factor we get a remainder. Similarly when we divide one polynomial by another which is not a multiple of  the other number we get a remainder and this is what the student learns about  the remainder theorem..

To make it easier for the students, this section in our NCERT Solutions for Class 9 Mathematics Chapter 2  starts with the divisor as a monomial followed by a binomial and then moving on to polynomial. There is also a six step in depth explanation  for the students to comprehend the division of a polynomial to master the topic with ease.

 2.5 Factorisation of Polynomials

You will get to learn about the factor theorem in detail in this section. By application of the remainder theorem you will learn to prove the factor theorem. You will come across various examples  of factor theorems which will  strengthen your factorizations of the polynomial and make you confident. . 

Factorisation of Polynomials requires  a good amount of practice and  students are recommended to refer to the  solved examples which are given in our NCERT Solutions for Class 9 Mathematics Chapter 2.

2.6 Algebraic identities 

In the earlier  classes, you have already come across various algebraic identities. In this section, you will learn how to factorise algebraic expressions using algebraic identities. You will also get to know their utilities in  applications. 

Furthermore you will learn about some advanced identities in the cubic form and the tips and tricks to solve any cubic polynomial within seconds.

To summarise, this chapter covers different concepts of algebra required in higher  classes from basic to advanced.

NCERT Solutions for Class 9 Mathematics Chapter 2 Exercise & Answer Solutions

NCERT Solutions for Class 9 Mathematics Chapter 2 Polynomials is available on the Extramarks official website. You can find theoretical explanations and have step-by-step solutions to all questions from the NCERT textbook.  During the exams , students should leverage the solution guide to  thoroughly practice in text questions and exercises given in  the chapter.

The solution guide on Extramarks website for the Class 9 Mathematics Chapter 2 covers topics on Polynomials in one variable, ways to apply factor theorem and remainder theorem and how to factorise a polynomial. Experts advise students to revise this  chapter multiple times as it’s a core topic in Algebra that helps define many equations and expressions. Algebraic equations goes a long way in the advanced level Mathematics syllabus for engineering, architecture and other science courses.

Click on the below links to view exercise specific questions and  exercises from  NCERT Solutions for Class 9 Mathematics Chapter 2:

  • Chapter 2: Exercise 2.1 – Question and Answer Solutions
  • Chapter 2: Exercise 2.2 – Question and Answer Solutions
  • Chapter 2: Exercise 2.3 – Question and Answer Solutions
  • Chapter 2: Exercise 2.4 – Question and Answer Solutions
  • Chapter 2: Exercise 2.5 – Question and Answer Solutions
  • Chapter 2: Exercise 2.6 – Question and Answer Solutions
  • Chapter 2: Exercise 2.7 – Question and Answer Solutions

Along with Class 9 Mathematics solutions, you can explore NCERT solutions on our Extramarks website for all primary and secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2, 
  • NCERT Solutions Class 3, 
  • NCERT Solutions Class 4, 
  • NCERT Solutions Class 5, 
  • NCERT Solutions Class 6, 
  • NCERT Solutions Class 7, 
  • NCERT Solutions Class 8, 
  • NCERT Solutions Class 9,
  • NCERT solutions Class 10,
  • NCERT solutions Class 11 
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar Class 9 Mathematics  is a complete source of information for CBSE students preparing for their  Class 9  exams. It consists of various examples with a detailed solution set. While revising from NCERT Exemplars, students  enhance their knowledge on the Mathematical  concepts and gain deeper insights on a variety of interconnected  topics between various chapters. 

Exemplars have not only proved beneficial for students in CBSE but also in other curriculum. It covers complex theories and problems that are easy to understand so that students can confidently prepare for  the upcoming examinations. Students appearing for Class 9 are advised to access  the NCERT Exemplar Class 9 Mathematics  and use it as their core study material. Students need not browse through other apps since Extramarks has built that credibility and is trusted by lakhs of students across India. After referring to the NCERT Solutions and NCERT Exemplar, the students can think logically about a problem. As a result, students can easily turn  to more advanced and higher-level conceptual questions.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 2

 To get good grades, it is necessary that you have well-structured notes which will help  you to recall and revise quickly whatever you have learnt. Therefore, NCERT Solutions for Class 9 Mathematics Chapter 2 offers a complete solution for all problems. The key features of NCERT Solutions   are: 

  • Extramarks NCERT Solutions for Class 9 Mathematics Chapter 2 has a point-to-point designed conceptual notes for ready reference.
  • The core concepts covered in Extramarks NCERT Solutions Class 9 Chapter 2 will help students to solve their examination papers in a more  logical manner and use their mathematical skills to answer any question.
  • With the  polynomial chapter solutions, students will be able to apply the concepts  required to solve many  algebraic  expressions.

Q.1 Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

i 4x23x+7ii y2+2iii 3t+t2iv y+2yvx10+y3+t50

Ans.

(i)   4x23x+7It is a polynomial in variable x.(ii)  y2+2 It is a polynomial in variable y.(iii)3t+t2=3t12+t2It is not a polynomial in variable t because power of t is in fraction i.e., 12.(iv)y+2y=y+2y1It is not a polynomial in variable y because power of y is negative i.e., 1.(v)x10+y3+t50It is not a polynomial in one variable because there are three variables i.e., x, y and t.

Q.2

Write the coefficients of x2 in each of the following:i 2+x2+xii 2x2+x3iiiπ2x2+xiv2x1

Ans.

(i)2+x2+xCoefficient of x2 is 1.(ii)2x2+x3Coefficient of x2 is  1.(iii)π2x2+xCoefficient of x2 is π2.(iv)2x1Coefficient of x2 is 0.

Q.3 Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Ans.

Binomial of degree 35 is x35 + 15.

Monomial of degree 100 is x100.

Q.4

Write the degree of each of the following polynomials:i 5x3+4x2+7xii 4y2 iii 5t7iv 3

Ans.

(i) 5x3+4x2+7x,The degree of the given polynomial is 3 becuase the heighest power of x is 3.(ii) 4y2, The degree of the given polynomial is 2 becuase heighest power of y is 2.(iii) 5t7,The degree of the given polynomial is 1 becuase heighest power of t is 1.(iv) 3The degree of the given polynomial is 0 becuase of constant term.

Q.5

Classifythefollowingaslinear,quadraticandcubicpolynomials:(i)x2+x    (ii)xx3(iii)y+y2+4(iv)1+x(v)3t(vi)r2(vii)7x3

Ans.

(i)x2+xThis polynomial is quadratic polynomial because  degree of the polynomial is 2.(ii)xx3This polynomial is cubic polynomial because  degree of the polynomial is 3.(iii)y+y2+4This polynomial is quadratic polynomial because  degree of the polynomial is 2.(iv)1+xThis polynomial is linear polynomial because  degree of the polynomial is 1.(v)3tThis polynomial is linear polynomial because  degree of the polynomial is 1.(vi)r2This polynomial is quadratic polynomial because  degree of the polynomial is 2.(vii)7x3This polynomial is cubic polynomial because  its degree is 3.

Q.6 Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = –1
(iii) x = 2

Ans.

(i) Putting x = 0, we get
P(0) = 5(0) – 4(0)2 + 3
= 3
(ii) Putting x = –1, we get
P(–1) = 5(–1) – 4(–1)2 + 3
= – 6
(iii) Putting x = 2, we get
P(2) = 5(2) – 4(2)2 + 3
= – 3

Q.7 Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3

(iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1)

Ans.

(i) p(y) = y2 – y + 1

Putting y = 0, we get
p(0) = (0)2 – (0) + 1 = 1
Putting y = 1, we get
p(1) = (1)2 – (1) + 1 = 1
Putting y = 2, we get
p(2) = (2)2 – (2) + 1 = 3
(ii) p(t) = 2 + t + 2t2t3

Putting t = 0, we get
p(0) = 2 + (0) + 2(0)2 –(0)3 = 2
Putting t = 1, we get
p(1) = 2 + (1) + 2(1)2 –(1)3 = 4
Putting t = 2, we get
p(2) = 2 + (2) + 2(2)2 –(2)3 = 4

(iii) p(x) = x3

Putting x = 0, we get
p(0) = (0)3 = 0
Putting x = 1, we get
p(1) = (1)3 =1
Putting x = 2, we get
p(2) = (2)3 = 8
(iv) p(x) = (x – 1) (x + 1)

Putting x = 0, we get
p(0) = (0 – 1) (0 + 1) = –1
Putting x = 1, we get
p(1) = (1 – 1) (1 + 1)= 0
Putting x = 2, we get
p(2) = (2 – 1) (2 + 1)= 3

Q.8

Verify whether the following are zeroes of the polynomial, indicated against them.ipx=3x+1,x=13iipx=5xπ,x=45iiipx=x21,x=1,1ivpx=(x+1)(x2),x=1,2vpx=x2,x=0vipx=lx+m,x=mlviipx=3x21,x=13,23viiipx=2x+1,x=12

Ans.

ipx=3x+1,x=13Putting x=13 in p(x), we getp13=3×13+1  =0Since, remainder is zero. So, x=13 is a zero of p(x).ii p x=5xπ,x=45Putting x=45 in p(x), we getp45=5×45π    =4π0Since, remainder is not zero. So, x=45 is not a zero of p(x).(iii)p(x)=x21,x=1,1Putting x=1 in p(x), we get       p1=(1)21  =0Since, remainder is zero. So, x=1 is a zero of p(x).Putting x=1 in p(x), we get  p1=(1)21=0Since, remainder is zero. So, x=1 is a zero of p(x).(iv)p(x)=(x+1)(x2),x=1,2Putting x=1 in p(x), we get   p(1)=(1+1)(12)  =0Since, remainder is zero. So, x =1 is a zero of p(x).Putting x=2 in p(x), we get     p(2)=(2+1)(22)=0Since, remainder is zero. So, x=2 is a zero of p(x).(v)p(x)=x2,x=0Putting x=0 in p(x), we get      p0=(0)2     =0Since, remainder is zero. So, x=0 is a zero of p(x).vi p x=lx+m,x=mlPutting x=ml in p(x), we get   pml=lml+m            =0Since, remainder is zero. So, x=ml is a zero of p(x).viipx=3x21,x=13,23Putting x=13 in p(x), we get   p13=31321              =0Since, remainder is zero. So, x=13 is a zero of p(x).Putting x=23 in p(x), we get   p23=32321                =30Since, remainder is not zero. So, x=13 is not a zero of p(x).viiipx=2x+1,x=12Putting x =12 in p(x), we get        p12=212+1          =1+1          =20Since, remainder is not zero. So, x=12 is not a zero of p(x).

Q.9

Find the remainder when x3+3x2+3x+1 is divided byi x+1iix12iii x iv x+π v 5+2x

Ans.

(i)x3+3x2+3x+1 is divided by x+1x2+2x+1x+1x3+3x2+3x+1 ±x3±x2¯2x2+3x+1±2x2±2x ¯ x+1±x±1¯0¯Thus, the remainder is 0.Another method:Let p(x)=x3+3x2+3x+1 and the zero of x+1 is 1.So, p(1)=(1)3+3(1)2+3(1)+1=1+33+1=0Thus, the remainder is 0. (ii)x3+3x2+3x+1 is divided by x12x2+72x+194x12x3+3x2+3x+1±x312x2¯72x2+3x+1±72x274x ¯194x+1±194x198¯278¯Thus, the remainder is 278.Another method:Let p(x)=x3+3x2+3x+1 and the zero of x12 is 12.So, p(12)=(12)3+3(12)2+3(12)+1=18+34+32+1=1+6+12+88=278 Thus, the remainder is 27 8 . (iii)x3+3x2+3x+1 is divided by xx2+3x+3xx3+3x2+3x+1±x3 ¯3x2+3x+1±3x2¯ 3x+1±3x¯1¯ Thus, the remainder is 1. Another method: Let p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of x is 0. So, p( 0 )= ( 0 ) 3 + 3 ( 0 ) 2 + 3( 0 ) + 1 =0+0+0+1 =1 Thus, the remainder is 1. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaWaaSaaaeaacaaIYaGaaG4naaqaaiaaiIdaaaGaaeOlaaqaamaabmaabaGaamyAaiaadMgacaWGPbaacaGLOaGaayzkaaGaaGPaVlaaykW7caWG4bWaaWbaaSqabeaacaaIZaaaaGqabOGaa8hiaiabgUcaRiaa=bcacaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhacaWFGaGaey4kaSIaa8hiaiaaigdacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabccacaqG4baabaGaaCzcaiaaxMaacaWLjaGaaeiEamaaAeaabaGaamiEamaaCaaaleqabaGaaG4maaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccaWFGaGaey4kaSIaa8hiaiaaiodacaWG4bGaa8hiaiabgUcaRiaa=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bcacaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhacaWFGaGaey4kaSIaa8hiaiaaigdacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG6bGaaeyzaiaabkhacaqGVbGaaeiiaiaab+gacaqGMbGaaeiiaiaabIhacaqGGaGaaeyAaiaabohacaqGGaGaaGimaiaab6caaeaacaqGtbGaae4BaiaabYcacaqGGaGaaeiCamaabmaabaGaaGimaaGaayjkaiaawMcaaiabg2da9maabmaabaGaaGimaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaakiaa=bcacqGHRaWkcaWFGaGaaG4mamaabmaabaGaaGimaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4mamaabmaabaGaaGimaaGaayjkaiaawMcaaiaa=bcacqGHRaWkcaWFGaGaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGimaiabgUcaRiaaicdacqGHRaWkcaaIWaGaey4kaSIaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaGaaeymaiaab6caaaaa@26CC@ (iv)x3+3x2+3x+1 is divided by x+πx2+(3π)x+(33π+π2)x+πx3+3x2+3x+1±x3±πx2¯(3π)x2+3x+1±(3π)x2±π(3π)x¯(33π+π2)x+1±(33π+π2)x±π(33π+π2)¯13π+3π2π3¯ Thus, the remainder is ( 13π+3 π 2 π 3 ). Another method: Let p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of x+π is π. So, p( π )= ( π ) 3 + 3 ( π ) 2 + 3( π ) + 1 = π 3 +3 π 2 3π+1 =13π+3 π 2 π 3 Thus, the remainder is 13π+3 π 2 π 3 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaWaaeWaaeaacaaIXaGaeyOeI0IaaG4maGGaaiab=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b8aWjaabccacaqGPbGaae4CaiaabccacqGHsislcqWFapaCcaqGUaaabaGaae4uaiaab+gacaqGSaGaaeiiaiaabchadaqadaqaaiabgkHiTiab=b8aWbGaayjkaiaawMcaaiabg2da9maabmaabaGaeyOeI0Iae8hWdahacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaaaaOGaa4hiaiabgUcaRiaa+bcacaaIZaWaaeWaaeaacqGHsislcqWFapaCaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccaGFGaGaey4kaSIaa4hiaiaaiodadaqadaqaaiabgkHiTiab=b8aWbGaayjkaiaawMcaaiaa+bcacqGHRaWkcaGFGaGaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaeyOeI0Iae8hWda3aaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG4maiab=b8aWnaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacqWFapaCcqGHRaWkcaaIXaaabaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIXaGaeyOeI0IaaG4maiab=b8aWjabgUcaRiaaiodacqWFapaCdaahaaWcbeqaaiaaikdaaaGccqGHsislcqWFapaCdaahaaWcbeqaaiaaiodaaaaakeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGTbGaaeyyaiaabMgacaqGUbGaaeizaiaabwgacaqGYbGaaeiiaiaabMgacaqGZbGaaeiiaiaaigdacqGHsislcaaIZaGae8hWdaNaey4kaSIaaG4maiab=b8aWnaaCaaaleqabaGaaGOmaaaakiabgkHiTiab=b8aWnaaCaaaleqabaGaaG4maaaakiaab6caaaaa@1A (v)x3+3x2+3x+1 is divided by 5+2x or 2x+512x2+14x+782x+5x3+3x2+3x+1±x3±52x2¯12x2+3x+1±12x2±54x¯74x+1±74x±358¯278¯ Thus, the remainder is 27 8 . Another method: Let p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of 5+2x is 5 2 . So, p( 5 2 )= ( 5 2 ) 3 + 3 ( 5 2 ) 2 + 3( 5 2 ) + 1 = 125 8 + 75 4 15 2 +1 = 125+15060+8 8 = 27 8 Thus, the remainder is 27 8 . 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Q.10 Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Ans.

x 3 ax 2 + 6xa is divided by xa MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaqG4bWaaWbaaSqabeaacaqGZaaaaOGaai4eGiaabggacaqG4bWaaWbaaSqabeaacaqGYaaaaOGaey4kaSIaaeiiaiaabAdacaqG4bGaai4eGiaabggacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabccacaqG4bGaai4eGiaabggaaaa@529D x 2 +6 xa x 3 ax 2 + 6xa ± x 3 a x 2 ¯ 6xa ±6x6a ¯ 5a ¯


T
hus, the denominator is 5a.

Another method:Let p(x)=x3ax2+ 6xa and the zero of xa is a.So, p(a)=(a)3(a)3+ 6(a)a=5aThus, the remainder is 5a.

Q.11 Check whether 7 + 3x is a factor of 3x3 + 7x.

Ans.

Let p(x)=3x3+ 7x is divided by 7 + 3x i.e., x=73 if 7+3x=0 .So, p(73)=3(73)3+ 7(73)=3439493=3431479=4909Thus, the remainder is 4909.Therefore, 7 + 3x is not a factor of 3x3+ 7x.

Q.12

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 x3x22+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)=  x3x2(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   = 220Since, remainder is 22. So, (x + 1) is not a factor of the given polynomial.

Q.13 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 2x1 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+126+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 )=x3 i.e.,x=3. So, P( 3 )= ( 3 ) 3 4 ( 3 ) 2 +( 3 )+6 =2736+3+6=0 Since, remainder is 0. So, g( x ) is a factor of given polynomial P( x ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@1105@

Q.14

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 = kx22x+1iv px = kx23x+k

Ans.

(i) p(x)=x2+x+kSince, x1 is a factor of p(x)=x2+x+k.So, by remainder theorem,    p(1)=012+1+k=0k=2Thus, the value of k is 2.ii px=2x2+kx+2Since, x1 is a factor of px=2x2+kx+2.So, by remainder theorem,                        p(1)=02(1)2+k(1)+2=0k=22Thus, the value of k is (2+2).iii px=kx22x+1Since, x1 is a factor of px=kx22x+1.So, by remainder theorem,                     p(1)=0k(1)22(1)+1=0k=1+2Thus, the value of k is (21). iv px=kx23x+kSince, x1 is a factor of px=kx23x+k.So, by remainder theorem,                 p(1)=0k(1)23(1)+k=0     2k=3       k=32Thus, the value of k is 32.

Q.15 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.16 Factorise:
(i) x3 – 2x2x + 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 2x2x+2 is divided by x2x21x2x3 2x2x+2±x32x2¯x+2x±2¯0¯Since,Dividend=Divisor×Quotient+Remainderx3 2x2x+2=(x2)(x21)+0=(x2)(x+1)(x1)

(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 x5x2+2x+1x5x3 3x2 9x 5±x35x2¯2x2 9x 5±2x210x¯x5 ±x5¯0¯Since,Dividend=Divisor×Quotient+Remainderx3 3x2 9x 5=(x5)(x2+2x+1)+0=(x5)(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+1x2+12x+20x+1x3+ 13x2+ 32x+ 20±x3±x2¯ 12x2+32x+20±12x2±12x¯20x+2020x+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 y1 2y2+y+1y12y3+y2 2y 1±2y32y2¯y22y1±y2y¯y1±y1¯0¯Since,Dividend=Divisor×Quotient+Remainder2y3+y2 2y1=(y1)(2y2+y+1)+0=(y1)(y+1)(2y+1)

Q.17

Use suitable identities to find the following products:ix+4x+10iix+8x10iii3x+43x5ivy2+32y232v 32x3+2x

Ans.

(i)       (x+4)(x+10)=x2+4x+10x+40             =x2+14x+40(ii)       (x+8)(x10)=x210x+8x80             =x22x80(iii)   (3x+4)(3x5)=9x215x+12x20             =9x23x20iv (y2+32)(y232)=(y2)2(32)2[(ab)(a+b)=a2b2]             =y494(v)    (32x)(3+2x)=(3)2(2x)2[(ab)(a+b)=a2b2]             =94x2

Q.18 Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 9

Ans.

(i) 103 × 107 = (100 + 3)(100 + 7)
= 10000 + (3 + 7)100 + 21
= 10000 + 1000 + 21
= 11021

(ii) 95 × 96 = (100 – 5)( 100 – 4)
= 10000 – (5 + 4)100 + 20
= 10000 – 900 + 20
= 9120

(iii) 104 × 96 = (100 + 4)(100 – 4)
= 10000 – (4)2 [(a – b)(a + b) = a2 – b2 ]
= 10000 – 16
= 9984

Q.19

Factorise​ the following using appropriate identities:i 9x2+6xy+y2ii 4y24y+1iii x2y2100

Ans.

(i)9x2+6xy+y2=(3x)2+2(3x)y+y2    =(3x+y)2           [a2+2ab+b2=(a+b)2]   =(3x+y)(3x+y)(ii)    4y24y+1=(2y)22(2y)1+(1)2   =(2y1)(2y1)[(ab)(a+b)=a2b2](iii)         x2y2100=x2(y10)2   =(xy10)(x+y10)[(ab)(a+b)=a2b2]

Q.20

Expand each of the following, using suitable identities:ix+2y+4z2ii2xy+z2iii2x+3y+2z2iv3a7bc2vi2x+5y3z2vii14a12b+12

Ans.

i(x+2y+4z)2=x2+(2y)2+(4z)2+2x(2y)+2(2y)(4z)+2(4z)x  =x2+4y2+16z2+4xy+16yz+8zx[(a+b+c)2=a2+b2+c2+2ab+2bc+2ca]ii  (2xy+z)2=(2x)2+y2+z22(2x)y2yz+2z(2x)  =4x2+y2+z24xy2yz+4zx[(ab+c)2=a2+b2+c22ab2bc+2ca]iii(2x+3y+2z)2              ={(2x3y2z)}2              =(2x3y2z)2       =(2x)2+(3y)2+(2z)22(2x)(3y)+2(3y)(2z)2(2z)(2x)              =4x2+9y2+4z212xy+12yz8zx[(abc)2=a2+b2+c22ab+2bc2ca]iv(3a7bc)2  =(3a)2+(7b)2+(c)22(3a)(7b)+2(7b)c2c(3a)           =9a2+49b2+c242ab+14bc6ca[(abc)2=a2+b2+c22ab+2bc2ca] v(2x+5y3z)2={(2x5y+3z)}2=(2x5y+3z)2=(2x)2+25y2+9z22(2x)(5y)2(5y)(3z)+2(3z)(2x)=4x2+25y2+9z220xy30yz+12zx[(ab+c)2=a2+b2+c22ab2bc+2ca]vi[14a12b+1]2=(14a)2+(12b)2+12(14a)(12b)2(12b)(1)+2(1)(14a)=116a2+14b2+114abb+12a[(ab+c)2=a2+b2+c22ab2bc+2ca]

Q.21

Factorise:(i)4x2+9y2+16z2+12xy24yz16xzii2x2+y2+8z222xy+42yz8xz

Ans.

(i) 4x2+9y2+16z2+12xy24yz16xz=(2x)2+(3y)2+(4z)2+2(2x)(3y)2(3y)(4z)2(4z)(2x)=(2x+3y4z)2 [(x+yz)2=x2+y2+z2+2xy2yz2zx]=(2x+3y4z)(2x+3y4z)(ii) 2x2+y2+8z222xy+42yz8xz=(2x)2+(y)2+(22z)22(2x)(y)+2(y)(22z)2(22z)(2x)=(2xy22z)(2xy22z)[(xyz)2=x2+y2+z22xy+2yz2zx]

Q.22

Write​ the following cubes in expanded form:i2x+13ii2a-3b3iii32x+13ivx-23y3

Ans.

i      (2x+1)3=(2x)3+3(2x)1(2x+1)+(1)3         =8x3+6x(2x+1)+1[(a+b)3=a3+3ab(a+b)+b3]         =8x3+12x2+6x+1ii  (2a3b)3=(2a)33(2a)(3b)(2a3b)(3b)3         =8a318ab(2a3b)27b3[(ab)3=a33ab(ab)b3]         =8a336a2b+54ab227b3iii    [32x+1]3=(32x)3+3(32x)1(32x+1)+13         =278x3+92x(32x+1)+1[(a+b)3=a3+3ab(a+b)+b3]         =278x3+274x2+92x+1iv[x23y]3=(x)33(x)(23y)(x23y)(23y)3         =x32x2y+49xy2827y3

Q.23 Evaluate the following using suitable identities:

(i) (99)3 (ii) (102)3 (iii) (998)3

Ans.

(i)         (99)3=(1001)3    =(100)33(100).1(1001)13[(ab)3=a33ab(ab)b3]    =1000000300×991    =1000000297001    =970299(ii)    (102)3=(100+2)3    = (100)3+3(100)(2)(100+2)+23 [(a+b)3=a3+3ab(a+b)+b3]    =1000000+600×102+8    =1000000+61200+8    =1061208(iii) (998)3=(10002)3    =(1000)33(1000)(2)(10002)23 [(ab)3=a33ab(ab)b3]    =10000000006000×9988    =994011992

Q.24 Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – (1/216) –(9/2)p2 + (1/4)p

Ans.

( i ) 8a 3 + b 3 +12 a 2 b+6a b 2 = ( 2a ) 3 + b 3 +3 ( 2a ) 2 b+3( 2a ) b 2 = ( 2a+b ) 3 [ ( a+b ) 3 = a 3 +3 a 2 b+3a b 2 + b 3 ] =( 2a+b )( 2a+b )( 2a+b ) 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( ii ) 8 a 3 b 3 12a 2 b+ 6ab 2 = ( 2a ) 3 b 3 3 ( 2a ) 2 b+3( 2a ) b 2 = ( 2ab ) 3 [ ( ab ) 3 = a 3 3 a 2 b+3a b 2 b 3 ] =( 2ab )( 2ab )( 2ab ) ( iii ) 27 125a 3 135a+ 225a 2 = ( 3 ) 3 ( 5a ) 3 3 ( 3 ) 2 ( 5a )+3( 3 ) ( 5a ) 2 = ( 35a ) 3 [ ( ab ) 3 = a 3 3 a 2 b+3a b 2 b 3 ] =( 35a )( 35a )( 35a ) ( iv ) 64 a 3 27b 3 144a 2 b+10 8ab 2 = ( 4a ) 3 ( 3b ) 3 3 ( 4a ) 2 ( 3b )+3( 4a ) ( 3b ) 2 = ( 4a3b ) 3 [ ( ab ) 3 = a 3 3 a 2 b+3a b 2 b 3 ] =( 4a3b )( 4a3b )( 4a3b ) 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( v ) 27p 3 ( 1 216 ) ( 9 2 ) p 2 + ( 1 4 )p = ( 3p ) 3 ( 1 6 ) 3 3 ( 3p ) 2 ( 1 6 )+3( 3p ) ( 1 6 ) 2 = ( 3p 1 6 ) 3 [ ( ab ) 3 = a 3 3 a 2 b+3a b 2 b 3 ] =( 3p 1 6 )( 3p 1 6 )( 3p 1 6 ) 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Q.25 Verify:
(i) x3+ y3 = (x + y) (x2xy+ y2)
(ii) x3 y3 = (x – y) (x2 + xy + y2)

Ans.

(i) x3+ y3 = (x + y) (x2 – xy+ y2)
Putting x = 1, y = 1 in L.H.S., we get
L.H.S. = x3+ y3 = 13+ 13 = 2
R.H.S. = (x + y) (x2xy+ y2)
= (1 + 1) (12 1.1+ 12)
= 2(1 – 1 + 1)
= 2
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 1 and y = 1.
Again, putting x = 2 and y = 3, we get
L.H.S. = x3+ y3 = 23+ 33 = 35
R.H.S. = (x + y) (x2xy+ y2)
= (2 + 3) (22 – (2).(3)+ 32)
= 5(4 – 6 + 9)
= 35
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 2 and y = 3.
Again, putting x = 3 and y = 0, we get
L.H.S. = x3+ y3 = 33+ 03 = 27
R.H.S. = (x + y) (x2xy+ y2)
= (3 + 0) (32 – (3).(0)+ 02)
= 3(9 – 0 + 0)
= 27
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 3 and y = 0.
In this way, it is verified that
x3+ y3 = (x + y) (x2xy+ y2)
(ii) x3 y3 = (x – y) (x2 + xy + y2)
Putting x = 1, y = 2 in L.H.S., we get
L.H.S. = x3 – y3 = 13 – 13 = 0
R.H.S. = (x – y) (x2 + xy + y2)
= (1 1) (12 1.1+ 12)
= 0
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 1 and y = 1.
Again, putting x = 2 and y = 3, we get
L.H.S. = x3 y3 = 23 33 = 19
R.H.S. = (x y) (x2 + xy+ y2)
= (2 3) (22 + (2).(3)+ 32)
= 1(4 + 6 + 9)
= 19
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 2 and y = 3.
Again, putting x = 3 and y = 0, we get
L.H.S. = x3 y3 = 33 03 = 27
R.H.S. = (x – y) (x2 + xy+ y2)
= (3 + 0) (32 + (3).(0)+ 02)
= 3(9 + 0 + 0)= 27
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 3 and y = 0.
In this way, it is verified that
x3 – y3 = (x – y) (x2 + xy + y2)

Q.26 Factorise each of the following:
(i) 27y3+ 125z3
(ii) 64m3 – 343n3

Ans.

(i) 27y3+ 125z3= (3y)3 + (5z)3
= (3y + 5z){(3y)2 – (3y)(5z) +(5z)2}
= (3y + 5z)(9y2 – 15yz + 25z2)
[Since, a3 + b3 =(a + b)(a2 – ab + b2)]

(ii) 64m3 –343n3= (4m)3 + (7n)3
= (4m – 7n){(4m)2 + (4m)(7n) +(7n)2}
= (4m – 7n)(16m2 + 28mn + 49n2)
[Since, a3 – b3 =(a – b)(a2 + ab + b2)]

Q.27 Factorise: 27x3 + y3 + z3 – 9xyz

Ans.

Since,
a3 + b3 + c3 – 3abc
= (a + b +c)(a2 + b2 + c2 – ab – bc – ca)
So, 27x3 + y3 + z3 – 9xyz
= (3x)3 + y3 + z3 – 3(3x)yz
= (3x + y + z){(3x)2 + y2 + z2 –(3x)y – yz – z(3x)}
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3xz)

Q.28

Verify that:x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

Ans.

R.H.S.=12(x+y+z)[(xy)2+(yz)2+(zx)2]  =12(x+y+z)[x22xy+y2+y22yz+z2+z22zx+x2]  =12(x+y+z)(2x2+2y2+2z22xy2yz2zx)  =(x+y+z)(x2+y2+z2xyyzzx)  =x3+y3+z33xyz=L.H.S.Hence proved. OR x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]For verifying the above equation, we have to put differentvalues of x, y and z in both sides.Putting x=0,y=1 and z=2 in L.H.S., we getL.H.S.=x3+y3+z33xyz    =03+13+233(0)(1)(2)    =1+8=9Putting x=0,y=1 and z=2 in R.H.S., we getR.H.S.=12(x+y+z)[(xy)2+(yz)2+(zx)2]              =12(0+1+2)[(01)2+(12)2+(20)2]              =32×6=9Thus, L.H.S.=R.H.S.Hence, it is verified that given equation is true for x=0,y=1 and z=2.Putting x=2,y=5 and z=3 in L.H.S., we getL.H.S.=x3+y3+z33xyz    =23+53+333(2)(5)(3)    =8+125+2790=70Putting x=2,y=5 and z=3 in R.H.S., we getR.H.S.=12(x+y+z)[(xy)2+(yz)2+(zx)2]    =12(2+5+3)[(25)2+(53)2+(32)2]    =12×10×14=70Hence, it is verified that given equation is true for x=2,y=5and z=3.Putting x=11,y=8 and z=5 in L.H.S., we getL.H.S.=x3+y3+z33xyz    =(11)3+83+533(11)(8)(5)   =1331+512+1251320=648Putting x=11,y=8 and z=5 in R.H.S., we getR.H.S.=12(x+y+z)[(xy)2+(yz)2+(zx)2]             =12(11+8+5)[(118)2+(85)2+(511)2]              =12×24×54=648Thus, L.H.S.=R.H.S.Hence, it is verified that given equation is true for x=11,y=8 and z=5.Thus,it is verified thatx3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

Q.29 If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Ans.

Since,
a3 + b3 + c3 – 3abc
= (a + b +c)(a2 + b2 + c2 – ab – bc – ca)
So, x3 + y3 + z3 – 3xyz
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
Putting x + y + z = 0, we get
x3 + y3 + z3 – 3xyz
= (0)(x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 – 3xyz = 0

x3 + y3 + z3 = 3xyz Hence Proved.

Q.30 Without actually calculating the cubes, find the value of each of the following:
(i) (–12)3 + (7)3 + (5)3
(ii) (28)3 + (–15)3 + (–13)3

Ans.

(i) (12)3+ (7)3+ (5)3 Let, x=12, y=7 and z=5then, x+y+z=12+7+5=0Since,       x3+ y3+ z3 3xyz   = (x+y+z)(x2+y2+z2xyyzzx)Putting  x + y + z = 0, we get       x3+ y3+ z3 3xyz   = (0)(x2+ y2+ z2 xy yz zx)      x3+ y3+ z3 3xyz = 0             x3+ y3+ z3= 3xyz  (12)3+ (7)3+ (5)3=3(12)(7)(5)                           =1260 (ii) (28)3+ (15)3+ (13)3 Let, x=28, y=15 and z=13then, x+y+z=281513          =0Since,       x3+ y3+ z3 3xyz   = (x+y+z)(x2+y2+z2xyyzzx)Putting  x + y + z = 0, we get       x3+ y3+ z3 3xyz   = (0)(x2+ y2+ z2 xy yz zx)      x3+ y3+ z3 3xyz = 0             x3+ y3+ z3= 3xyz (28)3+ (15)3+ (13)3      =3(28)(15)(13)      =16380

Q.31

Give possible expressions for the length and breadth ofeach of the following rectangles, in which their areas aregiven:i Area : 25a2-35a+12ii Area : 35y2+13y-12

Ans.

(i)Area:25a235a+12=25a235a+12           =25a220a15a+12           =5a(5a4)3(5a4)           =(5a4)(5a3)So, the length of rectangle is (5a4) and breadth is (5a3).(ii)Area:35y2+13y12=35y2+28y15y12           =7y(5y+4)3(5y+4)           =(5y+4)(7y3)So, the length of rectangle is (7y3) units and breadth is(5y+4)units.

Q.32 What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2–12x
(ii) Volume: 12ky2 + 8ky – 20k

Ans.

(i) Volume=3x212x =3x(x4)Thus,   the possible length of cuboid=3   units breadth of cuboid=x  units height of cuboid=(x4)  units(ii) Volume=12ky2+ 8ky20k = 4k(3y2+2y5) = 4k(3y2+5y3y5) = 4k(3y+5)(y1)Thus,   the possible length of cuboid=3k  units breadth of cuboid=(3y+5)  units height of cuboid=(y1)  units

Q.33 Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Ans.

(i) p(x) = x + 5
For zeroes of p(x),
let p(x) = 0,
so, x + 5 = 0 ⇒ x = – 5
Thus, – 5 is the zero of the given polynomial.
(ii) p(x) = x – 5
For zeroes of p(x),
let p(x) = 0,
so, x – 5 = 0 ⇒ x = 5
Thus, 5 is the zero of the given polynomial.
(iii) p(x) = 2x + 5
For zeroes of p(x),
let p(x) = 0,
so, 2x + 5 = 0 ⇒ x = – (5/2)
Thus, – (5/2) is the zero of the given polynomial.
(iv) p(x) = 3x – 2
For zeroes of p(x),
let p(x) = 0,
so, 3x – 2 = 0 ⇒ x = 2/3
Thus, 2/3 is the zero of the given polynomial.
(v) p(x) = 3x
For zeroes of p(x),
let p(x) = 0,
so, 3x = 0 ⇒ x = 0/3 = 0
Thus, 0 is the zero of the given polynomial.
(vi) p(x) = ax, a ≠ 0
For zeroes of p(x),
let p(x) = 0,
so, ax = 0 ⇒ x = 0/a = 0
Thus, 0 is the zero of the given polynomial.
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
For zeroes of p(x),
let p(x) = 0,
so, cx + d = 0 ⇒ x = – (d/c)
Thus, – (d/c) is the zero of the given polynomial.

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