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5 Classify the following as linear, quadratic and cubic polynomials ∵ The degree of x 2 x is 2 ∴ It is a quadratic polynomial ∵ The degree of x – x 3 is 3 ∴ It is a cubic polynomial ∵ The degree of y y 2 4 is 2 ∴ It is a quadratic polynomial ∵ The degree of 1 x is 1= (x y)(x 2 y 2 2xy x 2 xy y 2) using identity, (a b) 2 = a 2 b 2 2 ab) = (x y) (3xy) Hence, one of the factor of given polynomial is 3xy Question 18 The coefficient of x in the expansion of (x 3) 3 is (a) 1 (b) 9 (c) 18 (d) 27 Solution (d) Now, (x 3) 3 = x 3 3 3 3x (3)(x 3) using identity, (a b) 3 = a 3 b 3Materials Required Drawing sheet Pencil Cellotape Coloured papers Cutter Ruler Prerequisite Knowledge Square and its area Rectangle and its area Theory A square is a quadrilateral whose all

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(x y)^3 identity class 9-Here, Right hand side = Left hand side which means that (a3) (a3) is an identity Using Activity Method In this method, the algebraic identity is verified geometrically by taking different values of a x and ySelina Concise Mathematics Part I Solutions for Class 9 Mathematics ICSE, 4 Expansion All the solutions of Expansion Mathematics explained in detail by experts to help students prepare for their ICSE exams

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CBSE Class 9 Maths Lab Manual – Algebraic Identity (a – b) 2 = a 2 – 2ab b 2 Objective To verify the identity (a – b) 2 = (a 2 – 2ab b 2) by paper cutting and pasting Prerequisite Knowledge Area of a square = (side) 2 Area of a rectangle = l x b Materials Required A white sheet of paper, glazed papers, a pair of scissorsCBSE NCERT Notes Class 9 Maths Polynomials Show Topics Class 9 Maths Polynomials Algebraic Identities Algebraic Identities Algebraic identity is an algebraic equation that is true for all values of the variables occurring in it ( x y) 2 = x2 2 xy y2 ( x – y) 2 = x2 – 2 xy y2 x2 – y2 = ( x y) ( x – y)Example 12 Factorise the following (i) 21x2y3 27x3y2 (ii) a3 – 4a2 12 – 3a (iii) 4x2 – x 25 (iv) 2 –9 9 y (v) x4 – 256 Solution (i) 21x2y3 27x3y2 = 3 ×

An algebraic identity is an equality that holds for any values of its variables For example, the identity ( x y) 2 = x 2 2 x y y 2 (xy)^2 = x^2 2xy y^2 (x y)2 = x2 2xyy2 holds for all values of x x x and y y y Since an identity holds for all values of its variables, it is possible to substitute instances of one side of theEx 25, 9 Verify (i) x3 y3 = (x y) (x2 – xy y2) Ex 25, 9 Verify (ii) x3 y3 = (x y) (x2 xy y2) LHS x3 y3 We know (x y)3 = x3 y3 3xy (x yNCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)²

Y 3 ×Polynomials Exercise 25 Part 1 Question 1 Use suitable identities to find the following products (i) `(x 4)(x 10)` Answer Given, `(x 4)(x 10)` We know that, `(x a)(x b) = x^2 (a b) xThis video shows how to evaluate using the identity '(xy)3=x3y33x2y3xy2' To view more Educational content, please visit https//wwwyoutubecom/appuseri

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Solution 5 (i) 2 x 2 x is a quadratic polynomial as its degree is 2 (ii) x x 3 is a cubic polynomial as its degree is 3 (iii) y y 2 4 is a quadratic polynomial as its degree is 2 (iv) 1 x is a linear polynomial as its degree is 1 (v) 3t is a linear polynomial as its degree is 1The algebraic identities for class 9 consist of identities of all the algebraic formulas and expressions You must have learned algebra formulas for class 9, which are mathematical rule expressed in symbols but the algebraic identities represent that the equation is true for all the values of the variables For example;NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (a – b)³

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Solution (3x– 4y) 3 is of the form Identity VII where a = 3x and b = 4y So we have, (3x – 4y) 3 = (3x) 3 – (4y) 3 – 3(3x)(4y)(3x – 4y) = 27x 3 – 64y 3 – 108x 2 y 144xy 2 Example 5 Factorize (x 3 8y 3 27z 3 – 18xyz) using standard algebraic identities Solution (x 3 8y 3 27z 3 – 18xyz)is of the form IdentityIdentities VIII Last updated at by Teachoo Identity VII is a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac) Lets take an example a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac)Ex 25, 13 If x y z = 0, show that x3 y3 z3 = 3xyz We know that x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz zx) Putting x y z = 0, x3 y3 z3 3xyz = (0) (x2 y2 z2 xy yz zx) x3 y3 z3 3xyz = 0 x3 y3 z3 = 3xyz Hence proved

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Transcript Ex 25, 2 Evaluate the following products without multiplying directly (iii) 104 96 104 96 = (100 4) (100 4) Using the identity (x y) (x y) = x2 y2 where x = 100 , y = 4 = (100)2 (4)2 = 16 = 9984 Ex 25, 3 Factorise the following using appropriate identities (i) 9x2 6xy y2 9x2 6xy y2 = 32 x2 6xy (y)2 = (3x)2 6xy (y)2 = (3x)2 2 (3x) (y) (y)2Volume of cuboid = I x b x h Materials Required A set of 56 cubes each has dimensions (1 x 1 x 1) cubic unit Cubes may be of wood, plastic, cardboard or thermocol Procedure To verify the identity a 3 b 3, we shall take a = 3 units and b = 1 unit Make an arrangement of 28 cubes such that we get a cube of 3 x 3 x 3 cubic units and aCBSE Class 9 Maths Lab Manual – Algebraic Identity (a b) 3 = a 3 b 3 3a 2 b 3ab 2 Objective To verify the identity (ab) 3 = a 3 b 3 3a 2 b 3ab 2 geometrically by using sets of unit cubes Prerequisite Knowledge Volume of a cube = (edge) 3 Volume of a cuboid = l x b x h

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Polynomials Class 9 Important Questions Find value of polynomial 2x 2 5x 1 at x = 3 Check whether x = 1/6Polynomial Identities When we have a sum (difference) of two or three numbers to power of 2 or 3 and we need to remove the brackets we use polynomial identities (short multiplication formulas) (x y) 2 = x 2 2xy y 2 (x y) 2 = x 2 2xy y 2 Example 1 If x = 10, y = 5a (10 5a) 2 = 10 2 2
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5a (5a) 2 = 100 100a 25a 2(x1) (x2) = x 2 3x 2

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Materials Required Acrylic sheets Adhesive/Adhesive tape Scissors Geometry Box Cutter Prerequisite Knowledge Concept of cuboid and its volume Concept of cube and its volume Theory CuboidWe shall use the identity xy x y = x 2y 2 Here By applying in identity we get Hence the value of is (iv) The given expression is We have So we can express and in the terms of 100 as We shall use the identity xy x y = x 2y 2 Here By applying in identityFree download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 21, 22, 23, 24 and 25 of Polynomials in PDF form Online sols &

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State whether the following statements are true or false Give reasons to justify your answers (a) The degree of polynomial 5 x5 6 x4 8 x2 is 4 (b) The algebraic expression is a polynomial (c) The polynomial is a quadratic trinomial Using the long division method, determine the remainder when the polynomial 4 x5 2 x4 x3 4 x2 7The perfect cube forms ( x y) 3 (xy)^3 (xy)3 and ( x − y) 3 ( xy)^3 (x −y)3 come up a lot in algebra We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often ( x y) 3 = x 3 3 x 2 y 3 x y 2 y 3 ( x − y) 3 = x 3 − 3 x 2 y 3This video shows how to evaluate using the identity '(xy)3=x3y33x2y3xy2'To view more Educational content, please visit https//wwwyoutubecom/appuserie

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Algebraic Identity means that the lefthand side of the equation is identical to the righthand side of the equation, and for all values of the variables Algebra Formulas for Class 9 (x9 y 3)/(x7 y 8) so that the answer has no negative exponents Expand the logarithm log x 2 y 3 zClass 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 43 Question 1 Find the cube of each of the following binomial expressions Solution Question 2 If a b = 10 and ab = 21, find the value of a 3 b 3 Solution a b = 10, ab = 21ML Aggarwal Solutions for Class 9 Maths Chapter 3 – Expansions are provided here to help students prepare and excel in their exams This chapter mainly deals with problems based on expansions Experts tutors have formulated the solutions in a step by step manner for students to grasp the concepts easily From the exam point of view, solving

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Ex 25 Ex 25, 1 Ex 25,2 Important Ex 25,3 ^2 = (– √2 x y 2 √2 z )(– √2 x y 2 √2 z )In the 4th chapter of Class 9 RD Sharma Solutions students will study important identities as listed below Algebraic Identities Introduction Identity for the square of a trinomial Sum and difference of cubes Identity These books are widely used by the students who wish to score high in board examsWe shall use the identity xy x y = x 2y 2 Here By applying in identity we get Hence the value of is (iv) The given expression is We have So we can express and in the terms of 100 as We shall use the identity xy x y = x 2y 2 Here By applying in identity

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Safety How works Test new features Press Copyright Contact us CreatorsAbout Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy &For example, in x 3 y 3 3xy(x y), the degree of the polynomial is 3 For a non zero constant polynomial, the degree is zero Apart from these, there are other types of polynomials such as 32x 3 x – 1 The degree is 3;

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Algebraic Identities Polynomials, Class 9, Mathematics EduRev Notes is made by best teachers of Class 9 This document is highly rated by Class 9 students and has been viewed timesAlgebraic Identities Of Polynomials Example Problems With Solutions Example 1 Expand each of the following Solution (i) We have, Example 2 Find the products (i) (2x 3y) (2x – 3y) Solution (i) We have, Example 3 Evaluate each ofHello Students in this video we are going to discuss our new excercise in Class 9 NCERT Maths polynomials chapter, We have already discussed till now Remaind

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Answer/Explanation Answer b Explaination (b), A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R 5 A relation S in the set of real numbers is defined as xSy ⇒ x – y √3 is an irrational number, then relation S is (a) reflexive (b) reflexive and symmetric (c) transitiveCBSE Class 9 Maths Lab Manual – Algebraic Identity (a 3 – b 3) = (a – b) (a 2 ab b 2) Objective To verify the identity a 3 – b 3 = (a – b)(a 2 ab b 2) geometrically by using sets of unit cubes Prerequisite Knowledge Volume of a cube = (Edge) 3 Volume of a cuboid = l x b x hNCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)³

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OBJECTIVE To verify the algebraic identity (ab)²Ex 25 Class 9 Maths Question 12 Verify that x 3 y 3 z 33xyz = (x y z)(x y) 2 (yz) 2 (zx) 2 Solution We have, x 3 y 3 z 3 – 3xyz = (x y z) x 2 y 2 z 2 – xy – yz – zx = (x y z)2x 2 2y 2 2z 22xy2yz 2zx = (x y z)x 2 x 2 y 2 y 2 z 2 z 22xy2yz2zx = (x y z)x 2 y 2 – 2xy yY = 3 ×

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– 3ab (a – b) Materials Required Geometry box Acrylic sheet Scissors Adhesive/Adhesive tape Cutter2x 2 Classification of Polynomials Based on the Degrees of Its Variable Linear polynomial A polynomial having degree one is known as linear polynomialWe already have an identity for (x y) 3 So, let's try to derive the identity x 3 y 3 using the identity for (x y) 3 Let's first try to understand this geometrically Let's join our cubes as shown above We arranged both cubes in such a way to convert it into a cube as shown above

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Class IX Repair parts and components to include kits, assemblies, and subassemblies (repairable or nonrepairable) required for maintenance support of all equipment Class X Material to support nonmilitary programs such as agriculture and economic development (not included in Classes I through IX)Q4 Expand each of the following, using suitable identities (i) (x 2y 4z) 2 (ii) (2x – y z) 2 (iii) (–2x 3y 2z) 2 (iv) (3a – 7b – c) 2 (v) (–2x– 3ab (a – b) OBJECTIVE To verify the algebraic identity (a – b)³

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D) x=8 and y=0 Answer b Clarification If two complex numbers are equal, then corresponding parts are equal ie real parts of both are equal and imaginary parts of both are equalEx 25,5 Chapter 2 Class 9 Polynomials Last updated at by Teachoo Next Ex 25,6→ Chapter 2 Class 9 Polynomials;OBJECTIVE To verify the algebraic identity (ab)³

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Mathematics NCERT Grade 9, Chapter 2 Polynomials Polynomials are a particular type of algebraic expressionStudents will also study the remainder theorem and factor theorem in this chapterSome more algebraic identities will be discussed in this chapterTheir use in factorisation and in evaluating some given expressions will also be discussed Different polynomials based on the number of1, x – 3, –3/2(y) x Trinomial A polynomial having three terms is known as trinomial Example x²Get FREE NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 25 We have created Step by Step solutions for Class 9 maths to help you to revise the complete Syllabus and Score More marks Using identity (xy) 3 =x 3y 3 3xy(xy) = 8a 327b 318ab(2a3b) = 8a 3 – 27 b 3 – 36a 2 b 54ab 2 =8a 3 – 36a 2 b 54ab 2

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Y (7y 9x) (Using ab ac = a (b c)) = 3x2y2 (7y 9x) (ii) a3 – 4a2 12 – 3a = a2 (a – 4) – 3a 12 = a2 (a – 4

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