CLASSIX MATHEMATICS ASSIGNMENT2 CHAPTER – 2 POLYNOMIALS SECTIONA 1 Write the degree of the given polynomials i) ( 2x 4 )3 ii) ( t3 4 ) ( t3 9 ) 2 Write the coefficient of x4 and x in 4x3 5x4 2x2 3 3 Find the zeroes of f(z)=z2 2z 4 Find the product using suitable identities (4 5x)(45x) 5 What Transcript Example 18 Factorize 49a2 70ab 25b2 49a2 70ab 25b2 = 72 a2 70ab 52 b2 = (7a)2 70ab (5b)2 = (7a)2 (5b)2 2(7a) (5b) Using Identity (x y)2 = x2 y2 2xy where x = 7a and y = 5b = (7a 5b)2 = (7a 5b) (7a 5b) Example 18 Factorize (ii) 25/4 x2 y2/9 25/4 x2 y2/9 = 5^2/2^2 x2 y2/3^2 = (5/2 )^2 (y/3)^2 Using identity (a b) (a b) = a2 b2NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (a – b)³ = a³ – b³ – 3ab (a – b) OBJECTIVE To verify the algebraic identity (a – b)³ = a³ – b³ – 3ab (a – b) Materials Required Geometry box Acrylic sheet Scissors Adhesive/Adhesive tape Cutter

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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 yVolume of cuboid = I x b x h;




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Hi everyonein this video, I tell you about " x^3y^3=(xy)(x^2xyy^2) "?channel link https//wwwyoutubecom/channel/UC7Uui8og_cIpQaH9ItVWM3Q`````CBSE 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 cube = (edge) 3;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 7
NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)³ = a³b³ 3a²b 3ab² OBJECTIVE To verify the algebraic identity (ab)³ = a³b³ 3a²b 3ab² 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 CuboidDon't stop learning now Participate in the Scholorship Test for FirstSteptoDSA Course for Class 9 to 12 students My Personal Notes arrow_drop_upCBSE 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 scissors
Ex 25, 11 Factorise 27 𝑥3 𝑦3 𝑧3 – 9xyz 27 𝑥3 𝑦3 𝑧3 – 9𝑥𝑦𝑧 = 3𝑥3 𝑦3 𝑧3−9𝑥𝑦𝑧 = 3𝑥3 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 Identity Use the identity (x a) (x b) = x 2 (a b) x ab to find the following = x 2 y 2 z 2 – 6xyz 8 Question 3 Find the following squares by using the identities (i Attention reader!




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Hello 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 Ex 25, 1Use suitable identities to find the following products(i) (x 4) (x 10)(x 4) (x 10)Using (x a) (x b) = x2 (a b) x ab,Materials Required A set of 56 cubes each has dimensions (1 x 1 x 1) cubic unit




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Students can download RD Sharma Class 9 Maths Chapter 4 exercise 44 PDF from below link Chapter 4 Algebraic Identities Rd Sharma Class 9 Solutions Maths Chapter 1 Number System Rd Sharma Class 9 Solutions Maths Chapter 2 Exponents Of Real Numbers Rd Sharma Class 9 Solutions Maths Chapter 3 RationalisationPolynomial 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 2·10·5a (5a) 2 = 100 100a 25a 2 This 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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The 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 3Phone support is available MondayFriday, 900AM1000PM ET You may speak with a member of our customer support team by calling End of Conversation Have a great day! Therefore, by using the identity (xy) 2 = x 22xyy 2 p 2 –10p25 = (p5) 2 (iii) 25m 2 30m9 Ans Given 25m 2 30m9 Since, 25m 2 , 30m and 9 can be substituted by (5m) 2, 2×5m×3 and 3 2 respectively we get, = (5m) 2 2×5m×3 3 2 Therefore, by using the identity (xy) 2 = x 2 2xyy 2 25m 2 30m9 = (5m3) 2 (iv) 49y 2 84yz36z 2 Ans Given 49y 2




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= (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 3Rs Aggarwal 19 Solutions for Class 9 Math Chapter 3 Factorisation Of Polynomials are provided here with simple stepbystep explanations These solutions for Factorisation Of Polynomials are extremely popular among Class 9 students for Math Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for 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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Evaluate the following using suitable identities (i) (99) 3 (ii) (102) 3 (iii) (998) 3 Solution (i) (99) 3 = (100 – 1) 3 Identity (x – y) 3 = x 3 – y 3Ex 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 y(i) To prove x 3 y 3 = (x y) (x 2 – xy y 2) Now, using identity VI we can say (x y) 3 = x 3 y 3 3xy (xy) Or, (xy) 3 – 3xy (x y) = x 3 y 3




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For today's students, learning can happen anytime, anywhere Our mission is to improve 6th to 10th outcomes for all students and make learning more intuitive, more interesting, more personalised and more affordableUsing the above identity, the equation x 6 − y 6 can be factorised as follows x 6 Class 9 Maths Chapter 3 Unit 2 163 Qs > Related questions Factorize the following expression 8 x 3Class 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)




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Polynomials Class 9 Extra Questions Very Short Answer Type Question 1 Factorise 125x 3 – 64y 3 Solution 125x 3 – 6443 = (5x) 3 – (4y) 3 By using a3 – b3 = (a – b) (a 2 ab b 2), we obtain 125x 3 – 64y 3 = (5x – 4y) (25x 2 xy 16y 2) Question 2 Find the value of (x y) 2 (x – y) 2 Solution (x y) 2 (x – y) 2 = x 2 y 2 2xy x 2 y 2 – 2xy Since x − y = 3 xy=3 x − y = 3 implies y = x − 3, y=x3, y = x − 3, substituting this into the given identity gives a x (x − 3) b x c (x − 3) 9 = 0 a x 2 (− 3 a b c) x − 3 (c − 3) = 0 \begin{aligned} ax(x3)bxc(x3)9&=0\\ ax^2(3abc)x3(c3)&=0 \end{aligned} a x (x − 3) b x c (x − 3) 9 a x 2 (− 3 a b c) x − 3 (c − 3) = 0 = 0 9 (x y) 3 = x 3 y 3 3xy (x y) = x 3 3x 2 y 3xy 2 y 3 10 (x y) 3 = x 3 y 3 3xy (x y) = x 3 3x 2 y 3xy 2 y 3 11 x 3 y 3 z 3 3xyz = (x y z) (x 2 y 2 z 2 xy yz zx) 12 x 2 y 2 = ½ (x y) 2 (x – y) 2 13 xy = ¼ (x y) 2 (x – y) 2 14 x 2 y 2 = (x y




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Problem Solve (x 3) (x – 3) using algebraic identities Solution By the algebraic identity, x 2 – y 2 = (x y) (x – y), we can write the given expression as;By using standard formulae, expand the following (1 to 9) 1 (i) (2x 7y) 2 (ii) (1/2 x 2/3 y) 2 Solution (i) (2x 7y) 2 It can be written as = (2x) 2 2 × 2x × 7y (7y) 2 So we get = 4x 2 28xy 49y 2 (ii) (1/2 x 2/3 y) 2 It can be written as = (1/2 x) 2 2 × ½x 2/3y (2/3 y) 2 So we get = ¼ x 2 2/3 xy 4/9 y 2 2 (i) (3x 1/2x) 2 (ii) (3x 2 y 5z) 2Algebraic Identities Cubic Type ,Polynomials Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 9 on TopperLearning




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CBSE 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 Ex 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 yClass 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 = 21




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Transcript 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 provedThis video shows how to evaluate using the identity '(xy)3=x3y33x2y3xy2' To view more Educational content, please visit https//wwwyoutubecom/appuseriSelina Concise Mathematics Part I Solutions for Class 9 Mathematics ICSE, 5 Factorisation All the solutions of Factorisation Mathematics explained in detail by experts to help students prepare for their ICSE exams




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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) ( x a) ( x b) = x2 ( a b) x ab (x y z) 2 = x 2 y 2 z 2 2xy 2yz 2zxNCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)² = a² 2abb² OBJECTIVE To verify the algebraic identity (ab)² = a² 2abb² Materials 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 allAlgebraic Identities Polynomials, Class 9, Mathematics EduRev Notes is made by best teachers of Class 9 This document is highly rated by Class 9




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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 exams(9) Verify (i) x 3 y 3 = (x y) (x 2 − xy y 2) (ii) x 3 – y 3 = (x − y) (x 2 xy y 2) using some nonzero positive integers and check by actual multiplication Can you call these as identites ?We 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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(x 3) (x – 3) = x 2 – 3 2 = x 2 – 9 Problem Solve (x 5) 3 using algebraic identities Solution We know, (x y) 3 = x 3 y 3 3xy(xy) Therefore, (x 5) 3 = x 3 5 3 3x5(x5)




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