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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Polynomials Exercise 2.1 – 2.5

        Exercise 2.1

        Q.1     Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
                      (i) 4x2−3x+7        
                   
        (ii) y2+2–√
                      (iii) 3t√+t2–√   
                   
        (iv) y+2y
                      (v) x10+y3+t50
        Sol.

        (i) In 4x2−3x+7, all the indices of x are whole numbers so it is a polynomial in one variable x.
        (ii) In y2+2–√, the index of y is a whole number so it is a polynomial in one variable y.
        (iii) 3t√+t2–√=3t12+2–√t, here the exponent of the first term is 12, which is not a whole number.  Therefore it is not a polynomial.
        (iv) y+2y=y+2y−1, here the exponent of the second term is –1, which is not a whole number and so it is not a polynomial.
        (v) x10+y3+t50 is not a polynomial in one variable as three variables x, y, t occur in it.


        Q.2       Write the coefficients of x2 in each of the following :
                     
        (i) 2+x2+x                   
                     
        (ii) 2−x2+x3
                     
        (iii) π2x2+x   
                     
        (iv) 2x−−√−1
        Sol.

        Coefficient of x2:
        (i)  in2+x2+xis1
        (ii) in2−x2+x3is−1
        (iii) inπ2x2+xisπ2
        (iv) in2x−−√−1is0.


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

        Example :
        Binomial of degree 35 may be taken as x35+4x
        Monomial of degree 100 may be taken as 5x100


         

        Q.4       Write the degree of each of the following polynomials :
                     
        (i) 5x3+4x2+7x    
                     
        (ii) 4−y2
                     
        (iii) 5t−7–√                   
                     
        (iv) 3
        Sol.

        (i) The highest power term is 5x3 and the exponent is 3. So, the degree is 3.
        (ii) The highest power term is −y2 and the exponent is 2. So, the degree is 2.
        (iii) The highest power term is 5t and the exponent is 1. So, the degree is 1.
        (iv) The only term here is 3 which can be written as 3x0 and so the exponent is 0. Therefore the degree is 0.


        Q.5      Classify the following as linear, quadratic and cubic polynomials :
                    
        (i) x2+x    
                    
        (ii) x−x3     
                    
        (iii) y+y2+4
                    
        (iv) 1+ x                   
                    
        (v) 3t                         
                    
        (vi) r2
                    
        (vii) 7x3
        Sol.

        (i) The highest degree of x2+x is 2, so it is a quadratic polynomial .
        (ii) The highest degree of x−x3 is 3, so it is a cubic polynomial .
        (iii) The highest degree of y+y2+4 is 2, so it is a quadratic polynomial .
        (iv) The highest degree of x in (1 + x) is 1.  So it is a linear polynomial.
        (v) The highest degree of t in 3t is 1. So it is a linear polynomial.
        (vi) The highest degree of r in r2 is 2. So, it is a quadratic polynomial.
        (vii) The highest degree of x in 7x3 is 3. So, it is cubic polynomial.

        Exercise 2.2

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

        Let p(x)=5x−4x2+3
        (i)    At x = 0 : p(0)=5(0)−4(0)2+3
                 =0−0+3=3
        (ii)   At x = –1 : p(−1)=5(−1)−4(−1)2+3
                 =−5−4+3=−6
        (iii)  At x = 2 : p(2)=5(2)−4(2)2+3=10−16+3
                 =13−16=−3


        Q.2       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)

        Sol.

        (i)    We have , p(y)=y2−y+1
                 p(0)=(0)2−0+1=0−0+1=1
                 p(1)=(1)2−1+1=1−1+1=1
                 and p(2)=(2)2−2+1=4−2+1=3
        (ii)   We have p(t)=2+t+2t2−t3
                 Therefore p(0)=2+0+2(0)2−(0)3
                  =2+0+0−0=2
                  p(1)=2+1+2(1)2−(1)3
                  =2+1+2−1=5−1=4
                   and p(2)=2+2+2(2)2−(2)3
                   =2+2+8−8=4
        (iii)   We have, p(x)=x3
                   Therefore p(0)=(0)3=0
                   p(1)=(1)3=1
                   and p(2)=(2)3=8
        (iv)    We have p(x)=(x−1)(x+1)
                   Therefore  p(0)=(0−1)(0+1)=(−1)(1)=−1
                   p(1)=(1−1)(1+1)=(0)(2)=0
                   and p(2)=(2−1)(2+1)=(1)(3)=3


        Q.3        Verify whether the following are zeroes of the polynomial, indicated against them.
                       (i) p(x)=3x+1,x=−13

                           (ii) p(x)=5x−π,x=45
                           (iii) p(x)=x2−1,x=1,−1
                           (iv) p(x)=(x+1)(x−2),x=−1,2
                           (v) p(x)=x2,x=0
                           (vi) p(x)=lx+m,x=−mℓ
                           (vii) p(x)=3x2−1,x=−13√,23√
                           (viii) p(x)=2x+1,x=12
        Sol.

        (i)     We have , p(x) = 3x + 1
                  At x=−13, p(−13)=3(−13)+1=−1+1=0
                 Therefore −13 is a zero of polynomial = 3x + 1.
        (ii)    We have p(x)=5x−π
                  At  x=45
                  P(45)=5(45)−π=4−π
                  Therefore 45 is not a zero of polynomial 5x−π
        (iii)   We have, p(x)=x2−1
                   At x = 1, p(1)=(1)2−1=1−1=0
                   Therefore 1 is a zero of p (x).
                   Also , at x = -1 p(−1)=(−1)2−1
                   =1−1=0
                   Therefore -1 is a zero of polynomial x2−1
        (iv)     We have , p(x)=(x+1)(x−2)
                    At x = –1, p(−1)=(−1+1)(−1−2)
                    =(0)(−3)=0
                    Therefore -1 is  a zero of p (x) .
                    Also , at x = 2 , p(2)=(2+1)(2−2)=(3)(0)=0
                    Therefore 2 is zero of polynomial (x + 1)(x + 2).
        (v)       We have, p(x)=x2
                    At x = 0 , p(0)=(0)2=0
                    Therefore 0 is a zero of polynomial x2.
        (vi)      We have, p(x)=ℓx+m
                     At x=−mℓ p(−mℓ)=ℓ(−mℓ)+m
                     =−m+m=0
                     Therefore −mℓ is a zero of polynomial lx + m.
        (vii)     We have p(x)=3x2−1
                     At x=−13√ p(−13√)=3(13√)2−1
                     =3×−13−1=1−1=0
                     Therefore 13√ is a zero of polynomial 3x2−1
                     At x=23√ , p(23√)=3(23√)2−1=3×43−1
                     =4−1=3
                     Therefore 23√ is not a zero of polynomial 3x2−1
        (viii)    We have p(x)=2x+1
                      At x=12 p(12)=2(12)+1=1+1=2
                      Therefore 12 is not a zero of polynomial 2x + 1.


        Q.4        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.
        Sol.

        (i)     We have to solve p(x) = 0
                  ⇒ x + 5 = 0 ⇒ x = – 5
                  Therefore – 5 is a zero of the polynomial x + 5 .
        (ii)     We have to solve p(x) = 0
                   ⇒ x – 5 = 0 ⇒ x = 5
                   Therefore 5 is a zero of the polynomial x – 5.
        (iii)    We have to solve p(x) = 0
                    ⇒ 2x + 5 = 0 ⇒ x=−52
                    Therefore −52 is a zero of the polynomial 2x + 5.
        (iv)     We have to solve p(x) = 0
                    ⇒ 3x – 2 = 0 ⇒ x=23
                    Therefore 23 is a zero of the polynomial 3x – 2.
        (v)       We have to solve p(x) = 0
                    ⇒ 3x = 0 ⇒ x = 0
                    Therefore 0 is a zero of the polynomial 3x.
        (vi)      We have to solve p(x) = ax, a≠0
                     ⇒ ax = 0 ⇒ x = 0
                     Therefore 0 is a zero of the polynomial ax.
        (vii)     We have to solve p(x) = 0 , c≠0
                     ⇒ cx + d = 0 ⇒ x=−dc
                     Therefore −dc is a zero of the polynomial cx + d.

        Exercise 2.3

        Q.1      Find the remainder when x3+3x2+3x+1 is divided by
                        (i) x + 1                          (ii) x−12
                        (iii) x                              (iv) x+π            (v) 5 + 2x
        Sol.

        (i) By remainder theorem, the required remainder is equal to p (–1).
        Now, p(x)=x3+3x2+3x+1
        Therefore  p(−1)=(−1)3+3(−1)2+3(−1)+1
        =−1+3−3+1=0
        Hence, required remainder = p(–1) = 0

        (ii) By remainder theorem, the required remainder is equal to p(12).
        Now, p(12)=x3+3x2+3x+1
        =(12)3+3(12)2+3(12)+1
        =18+34+32+1=1+6+12+88
        =278
        Hence, required remainder =p(12)=278

        (iii) By remainder  theorem, the required remainder is equal to p (0).
        Now, p(x)=x3+3x2+3x+1
        Therefore  p(0)=0+0+0+1=1
        Hence, the required remainder = p(0) = 1

        (iv) By remainder theorem the required remainder is p(−π)
        Now , p(x)=x3+3x2+3x+1
        Therefore p(−π)=(−π)3+3(−π)2+3(−π)+1
        =−π3+3π2−3π+1
        Hence, the required remainder = p(x) =−π3+3π2−3π+1

        (v) By remainder theorem, the required remainder is p(−52)
        Now, p(x)=x3+3x2+3x+1
        Therefore p(−52)
        =(−52)3+3(−52)2+3(−52)+1
        =−1258+754−152+1
        =−125+150−60+88=−278
        Hence, required remainder =p(−52)=−278


        Q.2      Find the remainder when  x3−ax2+6x−a  is divided by x – a.
        Sol.

        Let p(x)=x3−ax2+6x−a
        By remainder theorem, when p(x) is divided by x – a.
        Then remainder = p(a)
        Therefore p(a)=a3−a.a2+6a−a
        =a3−a3+6a−a=5a
        Hence, required remainder = p(x) = 5a


         

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

        7 + 3x will be a factor of  p(x)=3x3+7xifp(−73)=0
        Now, p(−73)=3(−73)3+7(−73)
        =3×(−34327)−493=−3439−493≠0
        Therefore 7 + 3 x is not a factor  of  3x3+7x

         

        Exercise 2.4

        Q.1     Determine which of the following polynomials has (x + 1) a factor :
                      (i) x3+x2+x+1
                      (ii) x4+x3+x2+x+1
                      (iii) x4+3x3+3x2+x+1
                      (iv) x3+x2−(2+2–√)x+2–√ 
        Sol.

        (i) In order to prove that x + 1 is a factor of p(x)=x3+x2+x+1, it is sufficient to show that p(−1)=0
        Now, p(−1)=(−1)3+(−1)2+(−1)+1
        =−1+1−1+1=0
        Hence, (x + 1) is a factor of  p(x)=x3+x2+x+1

        (ii) In order to prove that (x + 1) is a factor of  p(x)=x4+x3+x2+x+1, it is sufficient to show that p (–1) = 0 .
        Now,  p(−1)=(−1)4+(−1)3+(−1)2+(−1)+1
        =1−1+1−1+1=1≠0
        Therefore (x + 1) is not a factor of   x4+x3+x2+x+1

        (iii) In order to prove that (x + 1) is a factor of p(x)=x4+3x3+3x2+x+1, it is sufficient to show that p(–1) = 0.
        Now, p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1
        =1−3+3−1+1=1≠0
        Therefore (x + 1) is not a factor of   x4+3x3+3x2+x+1.

        (iv) In order to prove that (x + 1) is a factor of  p(x)=x3−x2−(2+2–√)x+2–√,  it is sufficient to show that p(-1)= 0
        p(−1)=(−1)3−(−1)2−(2+2–√)(−1)+2–√
        =−1−1+2+2–√+2–√=22–√≠0
        Therefore   (x + 1) is not a factor of   x3−x2−(2+2–√)x+2–√


        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
        Sol.

        (i) In order to prove that g(x) = x + 1 is a factor of p(x)=2x3+x2−2x−1, it is sufficient to show that p(–1) = 0.
        Now, p(−1)=2(−1)3+(−1)2−2(−1)−1
        =−2+1+2−1=0
        Therefore g(x) is a factor of p(x).

        (ii) In order to prove that g(x) = x + 2 is a factor of  p(x)=x3+3x2+3x+1, it is sufficient to show that p(–2) = 0
        Now, p(−2)=(−2)3+3(−2)2+3(−2)+1
        =−8+12−6+1
        =−1≠0
        Therefore g(x) is not a factor of p(x) .

        (iii) In order to prove that g(x) = x – 3 is a factor of p(x)=x3−4x2+x+6. It is sufficient to show that p(+3) = 0
        Now,  p(3)=(3)3−4(3)2+3+6
        =27−36+3+6
        36 – 36 = 0
        Therefore g (x) is  a factor of p(x).


        Q.3      Find the value of k, if x -1 is a factor of p (x) in each of the following cases :
                    
        (i) p(x)=x2+x+k
                    
        (ii) p(x)=2x2+kx+2–√
                    
        (iii) p(x)=kx2−2–√x+1
                    
        (iv) p(x)=kx2−3x+k
        Sol.

        (i) If (x -1) is a factor of p(x)=x2+x+k, then
        p(1)=0
        ⇒ (1)2+1+k=0
        ⇒ 1+1+k=0
        ⇒ k=−2
        Hence, k = -2

        (ii) If (x – 1) is a factor of   p(x)=2x2+kx+2–√ , then
        p(1)=0
        ⇒ 2(1)2+k(1)+2–√=0
        ⇒ 2+k+2–√=0
        ⇒ k=−(2+2–√)

        (iii) If (x -1) is a factor of  p(x)=kx2−2–√x+1,  then
        p(1)=0
        ⇒ k(1)2−2–√(1)+1=0
        ⇒ k−2–√+1=0
        ⇒ k=2–√−1

        (iv) If (x -1) is a factor of p(x)=kx2−3x+k, then
        p(1)=0
        ⇒ k(1)2−3(1)+k=0
        ⇒ k−3+k=0
        ⇒ 2k=3
        ⇒ k=32


        Q.4      Factorise :
                       (i) 12x2−7x+1                            
                    (ii) 2x2+7x+3

                       (iii) 6x2+5x−6                          
                    (iv) 3x2−x−4

        Sol.

        (i) Here p + q = coeff. of x = – 7
        pq = coeff. of x2× constant term
        = 12 × 1 = 12
        Therefore p + q = – 7 = – 4 – 3
        and pq = 12 = (– 4)(–3)
        Therefore 12x2−7x+1=12x2−4x−3x+1
        =4x(3x−1)−1(3x−1)
        =(3x−1)(4x−1)

        (ii) Here p + q = coeff. of x = 7
        pq = coeff. of x2× constant term
        = 2 × 3 = 6
        Therefore p + q = 7 = 1 + 6
        and pq = 6 = 1 × 6
        Therefore 2x2+7x+3=2x2+x+6x+3
        =x(2x+1)+3(2x+1)
        =(2x+1)(x+3)

        (iii) Here p + q = coeff. of x = 5
        pq = coeff. of x2× constant term
        = 6 × (– 6)  = – 36
        Therefore p + q = 5 = 9 + (– 4)
        and pq = – 36 = 9 × (– 4)
        Therefore 6x2+5x−6=6x2+9x−4x−6
        =3x(2x+3)−2(2x+3)
        =(2x+3)(3x−2)

        (iv) Here p + q = coeff. of x = –1
        pq = coeff. of x2× constant term
        = 3 × (– 4) = – 12
        Therefore p + q = –1 = 3 + (– 4)
        and pq = – 12 = 3 × (–4)
        Therefore 3x2−x−4=3x2+3x−4x−4
        =3x(x+1)−4(x+1)
        =(x+1)(3x−4)


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

        Sol. 

        (i) Let f(x)=x3−2x2−x+2
        The constant term in f(x) is + 2 and factors of + 2 are ±1,±2
        Putting x = 1 in f(x),we have
        f(1)=(1)3−2(1)2−1+2
        =1−2−1+2=0
        Therefore (x–1) is a factor of f(x) .
        putting x = – 1 in f(x) , we have
        f(−1)=(−1)3−2(−1)2−(−1)+2
        =−1−2+1+2=0
        Therefore x + 1 is a factor of f (x).
        Putting x = 2 in f(x), we have
        f(2)=(2)3−2(2)2−(2)+2
        =8−8−2+2=0
        Therefore (x + 2) is a factor of f(x)
        Putting x = – 2 in f(x), we have
        f(−2)=(−2)3−2(−2)2−(−2)+2
        =−8−8+2+2
        =−12≠0
        Therefore x – 2 is not a factor of f (x).
        Therefore The factors of f(x) are (x – 1), (x + 1) and (x – 2).
        Let f(x) = k (x – 1) (x + 1) (x – 2)
        ⇒ x3−2x2−x+2=k(x−1)(x+1)(x−2)
        Putting x = 0 on both sides, we have
        2 = k (–1) (1) (–2) ⇒ k = 1
        Therefore x3−2x2−x+2=(x−1)(x+1)(x−2)

        (ii) Let p(x)=x3−3x2−9x−5
        We shall look for all factors of – 5. Therefore are ±1,±5.
        By trial, we find p (–1) = – 1 – 3 + 9 – 5 = 0 . So (x + 1) is a factor of p(x) .
        Now, divide p (x) by (x + 1)
        poly1
        Therefore p(x)=(x+1)(x2−4x−5)
        =(x+1)(x2+x−5x−5)
        =(x+1)[x(x+1)−5(x+1)]
        =(x+1)(x+1)(x−5)
        Therefore x3−2x2−x+2=(x+1)(x+1)(x−5)

        (iii) Let p(x)=x3+13x2+32x+20
        We shall look for all factors of + 20, these are ±1,±2,±4,±5,±10and±20
        By trial , we find
        p(−2)=(−2)3+13(2)2+32(−2)+20
        p(– 2) = – 8 + 52 – 64 + 20 = 0
        Therefore (x + 2) is a factor of p(x)
        Now, divide p (x) by x + 2
        poly2
        Therefore  p(x)=(x+2)(x2+11x+10)
        =(x+2)(x2+x+10x+10)
        =(x+2)[x(x+1)+10(x+1)]
        =(x+2)(x+1)(x+10)

        (iv) Let p(y)=2y3+y2−2y−1
        By trial we find p(1) = 2 + 1 – 2 – 1 = 0
        So, (y – 1) is a factor of p(y)
        Now, divide p(y) by y – 1
        poly3
        Therefore p(y)=(y−1)(2y2+3y+1)
        =(y−1)(2y2+2y+y+1)
        =(y−1)[2y(y+1)+1(y+1)]
        =(y−1)(y+1)(2y+1)

        Exercise 2.5

        Q.1     Use suitable identities to find the following products :
                      (i) (x + 4) (x + 10)                   (ii) (x + 8) (x – 10)
                   (iii) (3x + 4) (3x – 5)                (iv) (y2+32)(y2−32)
                   
        (v) (3 – 2x) (3 + 2x)
        Sol.

        (i) (x + 4) (x + 10) = x2+(4+10)x+4×10
                                         =x2+14x+40
        (ii) (x + 8) (x –10)  =x2+(8−10)x+8×(−10)
                                          =x2−2x−80
        (iii) (3x + 4) (3x – 5) = 3x(3x−5)+4(3x−5)
                                              =3x×3x−3x×5+4×3x−4×5
                                              =9x2−15x+12x−20
                                              =9x2−3x−20
        (iv) (y2+32)(y2−32)  =(y2)2−(32)2
                                                         =y4−94
        (v) (3 – 2x) (3 + 2x) = (3)2−(2x)2
                                            =9−4x2


        Q.2      Evaluate the following products without multiplying directly :
                    
        (i) 103 × 107      (ii) 95 × 96       (iii) 104 × 96
        Sol.

        (i) 103 × 107 = (100+3)(100+7)
                                 = (100)2+(3+7)(100)+3×7
                                 = 100×100+(10)(100)+21
                                 = 10000+1000+21=11021
        (ii) 95 × 96 = (100−5)(100−4)
                              = (100)2+(−5−4)(100)+(−5)(−4)
                              = 100×100+(−9)(100)+20
                               = 10000−900+20=9120
        (iii) 104 × 96 = (100+4)(100−4)
                                  = (100)2−(4)2
                                  = 10000−16=9984


        Q.3      Factorise the following using appropriate identifies :
                     
        (i) 9x2+6xy+y2                        
                     (ii) 4y2−4y+1
                     
        (iii) x2−y2100
        Sol.

        (a) 9x2+6xy+y2 = (3x)2+2(3x)(y)+(y)2 [(a+b)2=a2+2ab+b2]
                                                    = (3x+y)2=(3x+y)(3x+y)

        (b) 4y2−4y+1 = (2y)2−2(2y)(1)+(1)2 [(a−b)2=a2−2ab+b2]
                                            = (2y−1)2=(2y−1)(2y−1)
        (c) x2−y2100 = (x)2−(y10)2 [(x+y)(x−y)=x2−y2]
                                =(x−y10)(x+y10)


        Q.4      Expand each of the following using suitable identifies :
                    
        (i) (x+2y+4z)2
                    
        (ii) (2x−y+z)2
                    
        (iii) (−2x+3y+2z)2
                    
        (iv) (3a−7b−c)2
                    
        (v) (−2x+5y−3z)2
                    
        (vi) [14a−12b+1]2
        Sol.

        (i) (x+2y+4z)2
        =x2+(2y)2+(4z)2+2(x)(2y)+2(2y)(4z)+2(4z)(x)
        =x2+4y2+16z2+4xy+16yz+8zx

        (ii) (2x−y+z)2
        =[2x+(−y)+z]2
        =(2x)2+(−y)2+z2+2(2x)(−y)+2(−y)(z)+2(z)(2x)
        =4x2+y2+z2−4xy−2yz+4zx

        (iii) (−2x+3y+2z)2
        =[(−2x)+3y+2z]2
        =(−2x)2+(3y)2+(2z)2 +2(−2x)(3y)+2(3y)(2z)+2(2z)(−2x)
        =4x2+9y2+4z2−12xy+12yz−8zx

        (iv) (3a−7b−c)2
        =[3a+(−7b)+(−c)]2
        =(3a)2+(−7b)2+(−c)2+2(3a)(−7b)+2(−7b)(−c)+2(−c)(3a)
        =9a2+49b2+c2−42ab+14bc−6ca

        (v) (−2x+5y−3z)2
        =[(−2x)2+5y+(−3z)]2
        =(−2x)2+(5y)2+(−3z)2+2(−2x)(5y)+2(5y)(−3z)+2(−3z)(−2x)
        =4x2+25y2+9z2−20xy−30yz+12zx

        (vi) [14a−12b+1]2
        =[14a+(−12b)+1]2
        =(14a)2+(−12b)2+(1)2+2(14a)(−12b)+2(−12b)(1)+2(1)(14a)
        =116a2+14b2+1−14ab−b+12a


        Q.5      Factorise :
                       (i) 4x2+9y2+16z2+12xy−24yz−16xz
                       (ii) 2x2+y2+8z2−22–√xy−42–√yz−8xz
        Sol.

        (i) 4x2+9y2+16z2+12xy−24yz−16xz
        =(2x)2+(3y)2+(−4z)2+2(2x)(3y)+2(3y)(−4z)+2(2x)(−4z)
        = [2x+3y+(−4z)]2=(2x+3y−4z)2

        (ii) 2x2+y2+8z2−22–√xy+42–√yz−8xz
        =(2–√x)2+(−y)2+(−22–√z)2+2(2x−−√)(−y)+2(−y)(−22–√z)+2(2x−−√)(−22z−−√)
        =[2–√x+(−y)+(−22–√z)]2
        =(2–√x−y−22–√z)2


        Q.6      Write the following cubes in expanded form :
                       (i) (2x+1)3
                       (ii) (2a−3b)3
                       (iii) [32x+1]3
                       (iv) [x−23y]3
        Sol.

        (i) (2x+1)3 = (2x)3+3(2x)2(1)+3(2x)(1)2+(1)3 =8x3+12x2+6x+1

        (ii) (2a−3b)3 = (2a)3−3(2a)2(3b)+3(2a)(3b)2−(3b)3 = 8a3−36a2b+54ab2−27b3

        (iii)[32x+1]3=(32x)3+3(32x)2(1)+3(32x)(1)2+13 =278x3+274x2+92x+1

        (iv) [x−23y]3=x3−3(x)2(23y)+3(x)(23y)2−(23y)3 =x3−2x2y+43xy2−827y3


        Q.7      Evaluate the following using suitable identities
                        (i) (99)3
                        (ii) (102)3
                        (iii) (998)3
        Sol.

        (i) (99)3=(100−1)3
                   =(100)3−13−3(100)(1)(100−1)
                   =1000000−1−29700=970299

        (ii) (102)3=(100+2)3
        (100)3+(2)3+3(100)(2)(100+2)
        =1000000+8+61200=1061208

        (iii) (998)3=(1000−2)3
                       =(1000)3−(2)3−3(1000)(2)(1000−2)
                       =1000000000−8−5988000=994011992


        Q.8      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) 27p3−1216−92p2+14p
        Sol.

        (i) 8a3+b3+12a2b+6ab2
        =(2a)3+(b)3+3(2a)(b)(2a+b)
        =(2a+b)3
        =(2a+b)(2a+b)(2a+b)

        (ii) 8a3−b3−12a2b+6ab2
        =(2a)3−b3−3(2a)(b)(2a−b)
        =(2a−b)3
        =(2a−b)(2a−b)(2a−b)

        (iii) 27−125a3−135a+225a2
        =(3)3−(5a)3−3(3)(5a)(3−5a)
        =(3−5a)3
        =(3−5a)(3−5a)(3−5a)

        (iv) 64a3−27b3−144a2b+108ab2
        =(4a)3−(3b)3−3(4a)(3b)(4a−3b)
        =(4a−3b)3
        =(4a−3b)(4a−3b)(4a−3b)

        (v) 27p3−1216−92p2+14p
        =(3p)3−(16)3−3(3p)(16)(3p−16)
        =(3p−16)3=(3p−16)(3p−16)(3p−16)


        Q.9      Verify :
                     (i) x3+y3=(x+y)(x2−xy+y2)

                        (ii) x3−y3=(x−y)(x2+xy+y2)
        Sol.

        (i) R.H.S =(x+y)(x2−xy+y2)
                         =x(x2−xy+y2)+y(x2−xy+y2)
                         =x3−x2y+xy2+x2y−xy2+y3
                         =x3+y3=L.H.S.
        Thus, verified.
        (ii) R.H.S. =(x−y)(x2+xy+y2)
                           =x(x2+xy+y2)−y(x2+xy+y2)
                           =x3+x2y+xy2−x2y−xy2−y3
        x3−y3=L.H.S.
        Thus, verified.


        Q.10     Factorise each of the following :
                         (i) 27y3+125z3
                         (ii) 64m3−343n3
        Sol.

        (i) 27y3+125z3
        =(3y)3+(5z)3
        =(3y+5z)[(3y)2−(3y)5z+(5z)2]
        =(3y+5z)(9y2−15yz+25z2)

        (ii) 64m3−343n3 = (4m)3−(7n)3
                                  =(4m−7n)[(4m)2+(4m)(7n)+(7n)2]
                                  =(4m−7n)(16m2+28mn+49n2)


        Q.11     Factorise : 27x3+y3+z3−9xyz
        Sol.

        27x3+y3+z3−9xyz
        =(3x)3+y3+z3−3(3x)(y)(z)
        =(3x+y+z)[(3x)2+y2+z2−(3x)y−yz−z(3x)]
        =(3x+y+z)(9x2+y2+z2−3xy−yz−3zx)


         

        Q.12    Verify that x3+y3+z3−3xyz=12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
        Sol.

        R.H.S =12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
                   =12(x+y+z)(x2−2xy+y2+y2−2yz+z2+z2−2zx+x2)
                   =12(x+y+z)(2x2+2y2+2z2−2xy−2yz−2zx)
                   =22(x+y+z)(x2+y2+z2−xy−yz−zx)
                   =(x+y+z)(x2+y2+z2−yz−zx−xy−xy−yz−zx)
                   =x3+y3+z2−3xyz
                   = L.H.S.
        Hence verified.


        Q.13     If x + y + z = 0 , show that x3+y3+z3=3xyz.
        Sol.

        We have, x + y + z = 0
        ⇒ x + y = – z
        Cubing both sides, we have
        (x+y)3=(−z)3
        ⇒ x3+y3+3xy(x+y)=−z3
        ⇒ x3+y3−3xyz=−z3 [since x + y = – z]
        ⇒ x3+y3+z3=3xyz, which stands proved.


        Q.14     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
        Sol.

        (i) Let x = – 12, y = 7 and z = 5
        Here x+y+z=−12+7+5=0
        ⇒ x3+y3+z3=3xyz
        ⇒ (−12)3+(7)3+(5)3=3×(−12)×7×5=−1260

        (ii) Let x = 28, y = – 15 and z = – 13
        Here, x+y+z=28−15−13=0
        ⇒ x3+y3+z3=3xyz
        ⇒ (28)3+(−15)3+(−13)3
        =3(28)(−15)(−13)=16380


        Q.15     Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
                         (i) Area:25a2−35a+12   (ii) Area:35y2+13y−12
        Sol.

        Possible length and breadth of the rectangle are the factors of its given area.
        Area = length × breadth
        (i) Area:25a2+35a+12=25a2−15a−20a+12
                                                                   =5a(5a−3)−4(5a−3)=(5a−3)(5a−4)
        Therefore Possible length and breadth are (5a – 3) and (5a – 4) units.
        (ii) Area=35y2+13y−12
                            =35y2+28y−15y−12
                            =7y(5y+4)−3(5y+4)
                            =(5y+4)(7y−3)
        Therefore possible length and breadth are (5y + 4) and (7y – 3) units.


        Q.16     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
        Sol.

        Possible expressions for the dimensions of the cuboids are the factors of their volumes.
        (i) Volume =3x2−12x=3x(x−4)
        Therefore Possible dimensions of cuboid are 3, x and (x – 4) units

        (ii) Volume =12ky2+8ky−20k
                             =4k(3y2+2y−5)
                             =4k[3y(y−1)+5(y−1)]
                             =4k(y−1)(3y+5)
        Therefore Possible dimensions of cuboid are
        4k, (y – 1) and (3y + 5) units.

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