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How do I factorise 3(xy)^22(xy)?17 Factorisation (EMAG) Factorisation is the opposite process of expanding brackets For example, expanding brackets would require 2(x 1) to be written as 2x 2 Factorisation would be to start with 2x 2 and end up with 2(x 1) The two expressions 2(x 1) and 2x 2 are equivalent;How do you factor 2 x 2 5 x that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2BxC The values of r and s are equidistant from the center by an unknown quantity u

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Factorise 3(x y)^2-5(x y) 2 class 9-Quotient of x^38x^217x6 with x3;Q) x 3 3x 2 9x 5 Ans) p (x) = x 3 3x 2 9x 5 Factors of 5 = 1,5,1,5 p (1) = 1 3 3 x (1) 2 9 x (1) 5 = 1 3 9 5 = 1 17 = 16 ≠ 0 p (1) = (1) 3 x 3 x (1) 2 9 x (1) 5 = 1 3 9 5 = 9 9

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 Factorisation Class 8 Extra Questions Maths Chapter 14 Extra Questions for Class 8 Maths Chapter 14 Factorisation Factorisation Class 8 Extra Questions Very Short Answer Type Question 1 Find the common factors of the following terms (a) 25x2y, 30xy2 (b) 63m3n, 54mn4 Solution (a) 25x2y, 30xy2 25x2y = 5 × 5 × x × x  Ex 25 NCERT Solutions Polynomials Class 9 Notes EduRev is made by best teachers of Class 9 This document is highly rated by Class 9 students and has been viewed 3302 timesThe equation is in standard form 3x=5y2 3 x = 5 y 2 Divide both sides by 3 Divide both sides by 3 \frac {3x} {3}=\frac {5y2} {3} 3 3 x = 3 5 y 2 Dividing by 3 undoes the multiplication by 3 Dividing by 3 undoes the multiplication by 3

 So both x 2 − y 2 and − x y have x − y as a factor You can use these info to get below factorization x 2 y 2 − x y = ( x y) ( x − y) − ( x − y) = ( x y − 1) ( x − y)  Share answered Sep 7 '16 at 1523 Prince KumarExpand polynomial (x3)(x^35x2) GCD of x^42x^39x^246x16 with x^48x^325x^246x16;X^3 4x^2 6x 24 = 0 WolframAlpha Have a question about using WolframAlpha?

 NCERT Solutions Class 9 Maths Chapter 2 – Polynomials Exercise 25 are given here These NCERT Maths solutions are created by our subject experts which makes it easy for students to learn The students use it for reference while solving the exercise problems The fifth exercise in Polynomials Exercise 25 discusses the Algebraic Identities NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 25 Question 1 Use suitable identities to find the following products (i) (x 4) (x 10) (ii) (x8) (x 10) (iii) (3x 4) (3x – 5) (iv) (y2 ) (y2– ) (v) (3 – 2x) (3 2x) SolutionContact Pro Premium Expert Support »

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 Ex 25 Class 9 Maths Question 3 Factorise the following using appropriate identities (i) 9x 2 6xy y 2 (ii) 4y 2 – 4y 1 (iii) x 2 – Solution Ex 25 Class 9 Maths Question 4 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 5y – 3z) 2 (vi) 2 Solution Ex 25 Class 9 Maths Question 5 Example 14 Factorize y2 – 5y 6 by using the Factor Theorem Step 1 We check if y2 is multiplied by 1 y2 – 5y 6 Step 2 Writing y2 – 5y 6= (y – a) (y – b) So, ab = Constant term = 6 We find factors of 6 6 = 6 × 1 6 = 3 × 2 So, factors of 6 are 1, 2 and 3 We take value of y as 1 , 2, 3, –1, –2, –3, and check value of y2 – 5yView more examples » Access instant learning tools Get immediate feedback and guidance with stepbystep solutions and Wolfram

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POLYNOMIALS 31 Now observe the polynomials p(x) = 4x 5, q(y) = 2y, r(t) = t 2 and s(u) = 3 – uDo you see anything common among all of them?They have the same value for all values of x Factorise `x^(2)6xy^(2)9` days ago by Dhruvan Factorise `x^(2)6xy^(2)9` class9;

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The degree of each of these polynomials is one A polynomial of degree one is called a linear polynomial Some more linear polynomials in one variable are 2 x – 1, 2 y 1, 2 – uNow , try andNCERT Solution For Class 9 Maths Chapter 2 Polynomials 6 Write the following cubes in expanded form (i) (2x1)3 (ii) (2a−3b)3 (iii)( x1)3 (iv)(x− y)3Solutions Transcript Ex 25, 5 Factorise 4x2 9y2 16z2 12xy – 24yz – 16xz 4x2 9y2 16z2 12xy – 24yz – 16xz = 22 x2 32 y2 42 z2 12xy – 24yz – 16xz = (2x)2 (3y)2 (4z)2 12xy – 24yz – 16xz = (2x)2 (3y)2 (–4z)2 12xy – 24yz – 16xz = (2x)2 (3y)2 (–4z)2 2 (2x) (3y) 2 (3y) (–4z) 2 (2x) (–4z) Using (a b c)2 = a2 b2 c2

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Remainder of x^32x^25x7 divided by x3;= (3 x 5 y) (3 x – 5 y) Therefore, the factors of 9 x 2 – 25 y 2 are (3 x 5 y ) and (3 x – 5 y ) Factorising algebraic expression of form x 2 p x q edit Answer person Kishore Kumar Consider, 3 (x y) 2 – 5 (x y) 2 = 3 (x y) 2 – 3 (x y) – 2 (x y) 2 = 3 (x y) (x y) – 1 – 2 (x y) – 1 = (x y) – 1 3 (x y) – 2 = x y – 1 3x 3y – 2 Recommend (0) Comment (0)

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Factorise x 3 − 3 x 2 xAd by Grammarly Fast Answers Philip Macdonald Answered 1 year ago Author has 144 answers and 434K answer views You factor as follows Put a = (xy) Then 3(xy)^2 2(xy) becomes 3a^2 2a which is very easy to factor since a divides Class XII from Air Force Bal Bharti School Answered 3Exercise 25 Part 10 Question 9 – Verify (i) x3 y3 = (x y)(x2 − xy y2) x 3 y 3 = ( x y) ( x 2 x y y 2) Answer RHS = (x y)(x2 − xy y2 ( x y) ( x 2 x y y 2 = x3 − x2y xy2 yx2 − xy2 y3 = x 3 x 2 y x y 2 y x 2 x y 2 y 3 = x3 y3 = x 3 y 3 = LHS proved (ii) x3 y3 = (x− y)(x2 xy y2) x

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Telangana SCERT Class 9 Math Chapter 2 Polynomials and Factorisation Exercise 25 Math Problems and Solution Here in this Post Telanagana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 25Answer a 2 b 2 101 a 2 b 100 = a 2 b 2 100 a 2 b 1 a 2 b 100 = a 2 b a 2 b 100 1 a 2 b 100 = a 2 b 1 a 2 b 100 = a 2 b 1 a 2 b 100 Hence, factorisation of a 2 b 2 101 a 2 b 100 is a 2 b 1 a 2Evaluate (7/95/27) ×(1/35/18) class 7 Additive inverse of and reciprocal of 8 if y=a/ab , x=a/b what is yin terms of x A man puts by ` 5 in the first month and ` 2 more in every succeeding month

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 Factorise (xy)^3xy 2 See answers MsQueen MsQueen Hey mate! n1 I/K2 H>3J Q8 If the base of the triangle is 2xunits and its height is 5 units Thenfind the area of triangleNour answer Verify the property a × (b c) = a × b a × c for the followinga a = ( 3), b = 7, c = (9)b a = 4, b = 5, c = 7 Good evening!!!Come b gi lsFRANK Solutions Class 9 Maths Chapter 5 Factorisation 1 Factorise the following by taking out the common factors (a) 4x2y3 3– 6x y2 – 12xy2 (b) 5a (x 2 – y 2) 35b (x – y ) (c) 2x5y 8x3y2 2– 12x y3 (d) 12a3 2 15a b – 21ab2 (e) 24m4n6 6 56m n4 – 72m2n2 (f) (a – b)2 – 2(a – b) (g) 2a (p 2 q2) 4b (p q2) (h) 81 (p q)2 – 9p – 9q (i) (mx ny)2 2 (nx – my)

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7(x2y) – 25(x2y) 12 Solution Question 35 2(xy) – 9(xy) 5 Solution 2(xy) – 9(xy) 5 Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions Chapter 5 Exercise51 Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions Chapter 5 Exercise51 Q 1 Factorisation of Algebraic Expressions Exercise 51 I have a doubt regarding this question, kindly tell me whether the answer for the question given below is correct?Factorise a3xa2(xy)a(yz)z (x23x)25(x23x)y(x23x)5y x8y 8 Maths Polynomials This question has not been answered yet!

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 (iv) 3x 2x4 Solution Ex 24 Class 9 Maths Question 5 Factorise (i) x 32x 2x2 (ii) x 33x 29x5 (iii) x 313x 2 32x (iv) 2y 3 y 22y1 Solution (i) Let p(x) = x 32x 2x2, constant term of P(x) is 2 Factors of 2 are ± 1 and ± 2 Now, p(1)=1 32(1) 212 =1212 By trial we find that p(l) = 0, so (x 1) is a factor of p(x) SoIt is known that, (x y z) 2 = x 2 y 2 z 2 2xy 2yz 2zx 4x29y216z212xy–24yz–16xz =(2x)2(3y)2(−4z)2 2 (2x) (3y) 2 (3y) (−4z) 2 (−4z) (2x) =(2x3y–4z)2 =(2x3y–4z)(2x3y–4z) Concept Algebraic Identities Report Error Is there an error in this question or solution?In mathematics, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind For example, 3 × 5 is a factorization of the integer 15, and is a factorization of the polynomial x2 – 4 Factorization is not usually considered meaningful within number systems possessing division, such as the real or complex numbers, since any x {\displaystyle x

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Best solutions with stepbystep explanations and reasoning tips x2 −9 is a difference of squares and in general factorises as ∙ xa2 −b2 = (a −b)(a b) here a = x and b = 3 ⇒ x2 −9 = (x − 3)(x 3) Answer link Given a 4 2a 2 b 2 b 4 Since, a 4 and b 4 can be substituted by (a 2) 2 and (b 2) 2 respectively we get, = (a 2) 2 2×a 2 ×b 2 (b 2) 2 Therefore, by using the identity (xy) 2 = x 2 2xyy 2 a 4 2a 2 b 2 b 4 = (a 2 b 2) 2 Question 2 Factorise (i) 4p 2 –9q 2 (ii) 63a 2 –112b 2 (iii) 49x 2 –36 (iv) 16x 5 –144x 3 (v) (lm) 2(lm) 2 (vi) 9x 2 y 2 –16 (vii) (x 2 –2xyy 2

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Exercise 25 Polynomials Chapter 2 NCERT solution Class 8 NCERT text and video solutions that you will not find anywhere else!Using factor theorem, factorise the polynomial x3 x2 4x 4 0 votes 135k views asked in Class IX Maths by priya12 (12,184 points) Using factor theorem, factorise the polynomial x 3 x 2Exercise 51 Page No 59 Question 1 Factorize x 3 x – 3x 2 – 3 Solution x 3 x – 3x 2 – 3 Here x is common factor in x 3 x and – 3 is common factor in – 3x 2 – 3 x 3 – 3x 2 x – 3 x 2 (x – 3) 1(x – 3) Taking ( x – 3) common (x – 3) (x 2 1) Therefore x 3 x – 3x 2 – 3 = (x – 3) (x 2 1) Question 2 Factorize a(a b) 3 – 3a 2 b(a b)

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Factorise (i) 4x29y216z2 12xy−74yz−16xy (ii) 2x2 y28z2−2 2 y4 2 ,yz−8xz Write the following cubes in exprinded form (i) (2x1) ( 2 y)4 4 2 y2 w12 cubes in exprinded form (ii) (za−3b)2 (iii) {23 x13 (iv) x− 32 y Factorise (i)According to remainder theorem f (2) = 0 so that (x – 2 ) is a factor of x3 2x2 x 2 Here maximum power of x is 3 so that its can have maximum 3 factors So our answer is (x1) (x1) (x2) (ii) x3 3x2 9x 5 Possible zeros are factors of ± constant term / coefficient of leading term Here constant term is 5 and coefficient ofBest solutions with stepbystep explanations and reasoning tips

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 CBSE Class 9 Maths Polynomials Exercise 25 Algebraic Identities 1 (x y) 2 = x 2 2xy y 2 2 x 3 y 3 = (x y) (x 2 – xy y 2) (ii) x 3 – y 3 = (x – y) (x 2 xy y 2) (16m 2 49n 2 28mn) Q11 Factorise 27x 3 y 3 z 3 9xyz Answer ∵ x 3 y 3 z 3 3xyz = (x y z)(x 2 y 2 z 2 xy3x(x 2 2x – 5) Content Continues Below Factor 26 x sqrt9 y 3 – 13 y sqrt4 x 2 y xy sqrt169 x 4 y 3 , assuming that all variables are nonnegativeSelina solutions for Concise Mathematics Class 8 ICSE Chapter 13 FactorisationExercise 13 Page 159 Factorise `4/25 25b^2` Factorise 25 (2x y) 2 16 (x y) 2 Factorise ` (6 2/3)^2 (2 1/3)^2` Factorise ` (7 3/10)^2 (2 1/10)^2` Factorise 75 (x y) 2 48 (x y) 2 Factorise a 2 4a 4 b 2

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Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutorShare It On Facebook Twitter Email 1 Answer 0 votes answered by Lohith01 (970k points) selected by Dhruvan Best answer Given expression `x^(2)6xy^(2)9` Here, `x^(2)6x9` is perfect square Using formula, (x – y) 3 = x 3 – y 3 – 3xy(x – y) (99) 3 = (100 – 1) 3 = (100) 3 – 1 3 – (3 × 100 × 1) (100 – 1) = – 1 – 300(100 – 1) = – 1 – 300 = (ii) (102) 3 Solution 102 = 100 2 Using formula, (x y) 3 = x 3 y 3 3xy(x y) (100 2) 3 = (100) 3 2 3 (3 × 100 × 2) (100 2) = 8 600(100 2)

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