• Home
  • Courses
  • Online Test
  • Contact
    Have any question?
    +91-8287971571
    contact@dronstudy.com
    Login
    DronStudy
    • Home
    • Courses
    • Online Test
    • Contact

      Class 9 Maths

      • Home
      • All courses
      • 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

        Chapter Notes – Polynomials

        (1) Algebraic Expressions : Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
        For example: 5x, 2x – 3, x2 + 1, etc. are some algebraic expressions.


        (2) Polynomials : 
        The expression which contains one or more terms with non-zero coefficient is called a polynomial. A polynomial can have any number of terms.
        For example: 10, a + b, 7x + y + 5, w + x + y + z, etc. are some polynomials.


        (3) Polynomials in One Variable : The expression which contains only one type of variable in entire expression is called a polynomial in one variable.

        For example: 2x, a2 + 2a + 5, etc. are polynomials in one variable.


        (4) Term : A term is either a single number or variable and it can be combination of numbers and variable. They are usually separated by different operators like +, -, etc.

        For example: Consider an expression 6x – 7. Then, the terms in this expression are 6x and -7.


        (5) Coefficient : The number multiplied to variable is called as coefficient.
        For example: The coefficient of the term 2x will be 2.


        (6) Constant Polynomials : An expression consisting of only constants is called as constant polynomial.

        For Example: 7, -27, 3, etc. are some constant polynomials.


        (7) Zero Polynomial : The constant polynomial 0 is called as zero polynomial.

        (8) Denoting Polynomials in One Variable:
        Let us take an example to understand it:
        If the variable in a polynomial is x, then we can denote the polynomial by p(x) or q(x) etc.
        For example: p(x) = 7x2 + 7x + 7, t(r) = r3 + 2r + 1, etc.


        (9) Monomials : The expressions which have only one term are called as monomials.
        For Example: p(x) = 3x, q(a) = 2a2, etc. are some monomials.


        (10) Binomials : The expressions which have two terms are called as binomials.
        For example: r(x) = x + 10, c(z) = 7z2 + z etc. are some binomials.


        (11) Trinomials : 
        The expressions which have three terms are called as trinomials.
        For example: p(x) = 7x2 + x + 7, d(t) = t3 – 3t + 4, etc. are some trinomials.


        (12) Degree of polynomial : The highest power of the variable in a polynomial is called as the degree of the polynomial.
        For Example: The degree of p(x) = x5 – x3 + 7 is 5.
        Note: The degree of a non-zero constant polynomial is zero.


        (13) Linear polynomial : A polynomial of degree one is called a linear polynomial.
        For Example: 2x – 7, s + 5, etc. are some linear polynomials.


        (14) Quadratic polynomial : A polynomial having highest degree of two is called a quadratic polynomial. In general, a quadratic polynomial can be expressed in the form ax2 + bx + c, where a≠0 and a, b, c are constants.
        For Example: x2– 9, a2 + 7, etc. are some quadratic polynomials.


        (15) Cubic polynomial : A polynomial having highest degree of three is called a cubic polynomial. In general, a quadratic polynomial can be expressed in the form ax3 + bx2 + cx + d, where a≠0 and a, b, c, d are constants.
        For Example: x3– 9x +2, a3 + a2 + a + 7, etc. are some cubic polynomials.


        (16) General expression of polynomial : A polynomial in one variable x of degree n can be expressed as an xn + an-1 xn-1 + ….. + a1 x + a0, where an ≠ 0 and a0, a1, …. an are constants.

        (17) Zeroes of a Polynomial : The value of variable for which the polynomial becomes zero is called as the zeroes of the polynomial.
        For Example: Consider p(x) = x + 2. Find zeroes of this polynomial.
        (i) If we put x = -2 in p(x), we get,
        (ii) p(-2) = -2 + 2 = 0.
        (iii) Thus, -2 is a zero of the polynomial p(x).


        (18) Some Note-worthy Points:
        (i) A non-zero constant polynomial has no zero.
        (ii) A linear polynomial has one and only one zero.
        (iii) A zero of a polynomial might not be 0 or 0 might be a zero of a polynomial.
        (iv) A polynomial can have more than one zero.


        (19) Some Examples:
        For Example: Find value of polynomial 3a2 + 5a + 1 at a = 3.
        (i) Here, p(a) = 3a2 + 5a + 1.
        (ii) Now, substituting a = 3, we get,
        (iii) p(3) = 3 x (3)2 + 5 x 3 + 1 = 27 + 15 + 1 = 43

        For Example: Check whether at x = -1/7 is zero of the polynomial p(x) = 7x + 1.
        (i) Given, p(x) = 7x + 1.
        (ii) Now, substituting x = -1/7, we get,
        (iii) p(-1/7) = 7(-1/7) + 1 = -1 + 1 = 0.
        (iv) Here, p(-1/7) is zero. Thus, -1/7 is zero of the given polynomial.

        For Example: Find zero of the polynomial p(x) = 2x+ 2.
        (i) Equating p(x) to zero, we get,
        (ii) p(x) = 0
        (iii) 2x + 2 = 0
        (iv) 2x = -2 i.e. x = -1.
        (v) Thus, x = -1 is a zero of the given polynomial.

        (20) Remainder Theorem:
        Statement: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
        Proof :
        (i) Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x), i.e., p(x) = (x – a) q(x) + r(x)
        (ii) Since the degree of (x – a) is 1 and the degree of r(x) is less than the degree of (x – a), the degree of r(x) = 0. This means that r(x) is a constant, say r.
        (iii) So, for every value of x, r(x) = r.
        (iii) Therefore, p(x) = (x – a) q(x) + r
        (iv) In particular, if x = a, this equation gives us
        (v) p(a) = (a – a) q(a) + r = r, which proves the theorem.
        In other words, If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x)≠0, then there exists two polynomials q(x) and r(x) such that p(x) = g(x)q(x) + r(x), where, q(x) represents the quotient and r(x) represents remainder when p(x) is divided by g(x).

        For Example: Divide 3x2 + x – 1 by x + 1.
        (i) Let, p(x) = 3x2 + x – 1 and g(x) = x + 1.
        (ii) Performing divisions on these polynomials, we get,(iii) Now, we can re-write p(x) as 3x2 + x – 1 = (x + 1) (3x -2) +1.

        For Example: Find remainder on dividing x3 + 3x2 + 3x + 1 by 2x + 5.Thus, remainder obtained on dividing x3 + 3x2 + 3x + 1 by 2x + 5 is -27/8.

        (21) Factorisation of Polynomials:
        (i) Factor Theorem: If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then
        (a) x – a is a factor of p(x), if p(a) = 0
        (b) p(a) = 0, if x – a is a factor of p(x)

        For Example: Check whether (x + 1) is factor of p(x) = x3 + x2 + x + 1.
        (i) As per Factor Theorem, (x + 1) is factor of p(x) = x3 + x2 + x + 1, if p(-1) = 0.
        (ii) Therefore, p(-1) =(-1)3 + (-1)2 + (-1) + 1 = -1 + 1 -1 + 1 = 0.
        (iii) Thus, (x + 1) is factor of p(x) = x3 + x2 + x + 1.

        For Example: Find value of k, if (x – 1) is factor of p(x) = kx2 – 3x + k.
        (i) As per Factor theorem, here, p(1) = 0.
        (ii) So, k(1)2 – 3(1) + k = 0.
        (iii) k – 3 + k = 0
        (iv) 2k – 3 = 0
        (v) k = 3/2.

        For Example: Factorise 2y3 + y2 – 2y – 1.
        (i) On using trial and error method, we get,
        (ii) p(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2 + 1 – 2 -1 = 0.
        (iii) Thus, (y – 1) is factor of 2y3 + y2 – 2y – 1.
        (iv) Now, using division method, we get,(v) Thus, p(y) = 2y3 + y2 – 2y – 1
        = (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)


        (22) Algebraic Identities:
        (i) (a +b) 2 = (a2 + 2ab + b2)
        (ii) (a – b) 2 = (a2 – 2ab + b2)
        (iii) a2 – b2 = (a + b) (a – b)
        (iv) (x + a) (x + b) = x2 + (a + b)x + ab
        (v) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
        (vi) (a + b)3 = a3 + b3 + 3ab (a + b)
        (vii) (a – b) 3 = a3 – b3 – 3ab (a – b) = a3 – 3a2b + 3ab2 – b3
        (viii) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

        For Example: Use suitable identity to find (x + 2) (x – 3).
        (i) We know the identity, (x + a) (x + b) = x2 + (a + b)x + ab
        (ii) Using the identity, (x + 2) (x – 3) = x2 + (2 – 3)x + (2)(-3) = x2 – x – 6.

        For Example: Evaluate (102 x 107) without multiplying directly.
        We know the identity, (x + a) (x + b) = x2 + (a + b)x + ab
        (i) Here, we can write, 102 as (100 + 2) and 107 as (100 + 7). So, x = 100, a = 2 and b = 7.
        (ii) Using the identity, (102 x 107) = 1002 + (2 + 7)100 + (2)(7) = 10000 + 900 + 14 = 10914

        For Example: Factorise (a + b + c)2 = 4a2 + 16b2 + 64c2 + 16ab + 64bc + 32ca.
        (i) We know the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
        (ii) Now, 4a2 + 16b2 + 64c2 + 16ab + 64bc + 32ca
        = (2a)2 + (4b)2 + (8c)2 + 2(2a)(4b) + 2(4b)(8c) + 2(8c)(2a).
        = (2a + 4b + 8c)2
                     = (2a + 4b + 8c) (2a + 4b + 8c)

        For Example: Write (x – 2/3y)3 in expanded form.
        (i) We know the identity, (a – b) 3 = a3 – b3 – 3ab (a – b)
        (ii) (x – 2/3y)3 = x3 – (2/3y)3 – 3(x)(2/3y) (x – 2/3y)
        = x3 – 8/27y3 – 2xy (x – 2/3y)
        = x3 – 8/27y3 – 2x2y + 4/3 xy2

        For example: Factorise 8a3 + 27b3 + 64c3 – 72abc.
        (i) We know the identity, a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
        (ii) So, 8a3 + 27b3 + 64c3 – 72ab
        = (2a)3 + (3b)3 + (4c)3 – 3(2a)(3b)(4c)
        = (2a +3b + 4c) ((2a)2 + (3b)2 + (4c)2 – (2a)(3b) –(3b)(4c) -(4c)(2a))
        = (2a + 3b + 4c) (4a2 + 9b2 + 16c2 – 6ab – 12bc – 8ca)

        Prev Factorisation of Polynomial by Using Algebric Identities
        Next NCERT Solutions – Polynomials Exercise 2.1 – 2.5

        Leave A Reply Cancel reply

        Your email address will not be published. Required fields are marked *

        All Courses

        • Backend
        • Chemistry
        • Chemistry
        • Chemistry
        • Class 08
          • Maths
          • Science
        • Class 09
          • Maths
          • Science
          • Social Studies
        • Class 10
          • Maths
          • Science
          • Social Studies
        • Class 11
          • Chemistry
          • English
          • Maths
          • Physics
        • Class 12
          • Chemistry
          • English
          • Maths
          • Physics
        • CSS
        • English
        • English
        • Frontend
        • General
        • IT & Software
        • JEE Foundation (Class 9 & 10)
          • Chemistry
          • Physics
        • Maths
        • Maths
        • Maths
        • Maths
        • Maths
        • Photography
        • Physics
        • Physics
        • Physics
        • Programming Language
        • Science
        • Science
        • Science
        • Social Studies
        • Social Studies
        • Technology

        Latest Courses

        Class 8 Science

        Class 8 Science

        ₹8,000.00
        Class 8 Maths

        Class 8 Maths

        ₹8,000.00
        Class 9 Science

        Class 9 Science

        ₹10,000.00

        Contact Us

        +91-8287971571

        contact@dronstudy.com

        Company

        • About Us
        • Contact
        • Privacy Policy

        Links

        • Courses
        • Test Series

        Copyright © 2021 DronStudy Pvt. Ltd.

        Login with your site account

        Lost your password?

        Modal title

        Message modal