We get ,x 2y 2−z 2−2xy =(x−yz)(x−y−z)Factorisex2y2−z2−2xy =x2y2−z2−2xy=x2y2−2xy−z2=x−y2− Factorise x2y2−z2−2xy Please scroll down to see the correct answer and solution guideFactorise x^22xyy^2〖4z〗^2Factorise x2 2xy y2 4z2
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Factorise x^2+y^2+z^2-xy-yz-zx-Answer (1 of 2) \begin{array}{l}\text{Let E= } \\ x\left(y^{2}z^{2})y\left(y^{2}x^{2}\right)3\left(x^{2}y^{2}\right)2 x y z\right \\ Ex 142, 2 (vii) Chapter 14 Class 8 Factorisation Last updated at by Teachoo Solve all your doubts with Teachoo Black (new monthly pack available now!) Join Teachoo Black Next Ex 142, 2 (viii) Important → Chapter 14 Class 8
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Factorise (x 2−2xyy 2)−z 2 It is x2 −y2 z2 −2xz = (x2 −2xz z2) −y2 = (x − z)2 −y2 = (x −z y) ⋅ (x −z − y) Answer link//mathstackexchangecom/questions//parametrizeavarietyinprojectivespace You complete the square and you dont have to do the affine and infinity line separately, this is unnatural So complete the x, square = x^22 (yz)xy^2z^22yz= (xyz)^24yz So you have (xyz)^2=4yz
Factorise x2 y2 – z2 – 2xy Open in App Solution x 2 y 2z 22 x y = x 2 y 22 x yz 2 = xy 2z 2 a 22 a b b 2 = ab 2 = xy z xyz a 2b 2 = a b ab Suggest Corrections 0If in the general integer solution, the parameter n is replaced by −n, then one has {x,y} = {k(an2 2mn), k(m2 −n2)}, z = k(−amn−m2 −n2) This still describes all integer solutions, Just plug x2 2xy = 0 into your first equation You get y = 1 or y = −1 Then plug those two into the solutions you got for x So you get x = −2 or x = 0 or x = 2 The factorisation could be done by introducing 3 units, i, j, k such that, i 2 = j 2 = k 2 = 1 i j = j k = k i = − 1 which sort of look like quaternions, but not quite In this case the factorisation becomes, x 2 y 2 z 2 − 2 x y − 2 y z − 2 z x = ( i x j y k z) 2
Expand polynomial (x3)(x^35x2) GCD of x^42x^39x^246x16 with x^48x^325x^246x16Answer (1 of 2) The answer I see in Quora is not the factorisation of x^2y^2z^2xyyzzx —(1) It is merely a different way of representing (1) So it is absolute a wrong answer The true factorisation of (1) is as follows (1)=(xy*wz*w^2)(xy*w^2z*w) Where w is the complex cube root of uAlgebra Factor x^22xyy^2 x2 2xy y2 x 2 2 x y y 2 Check that the middle term is two times the product of the numbers being squared in the first term and third term 2xy = 2 ⋅x⋅y 2 x y = 2 ⋅ x ⋅ y Rewrite the polynomial x2 2⋅x⋅yy2 x 2 2 ⋅ x ⋅ y y 2
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Factorise x^2y^2z^22xy2xz2yz Brainlyin ImS44DFactoring » Tips for entering queries Enter your queries using plain English To avoid ambiguous queries, make sure to use parentheses where necessary Here are some examples illustrating how to ask about factoring factor quadratic x^27x12; See below Making m = xy n = yz p = xz we have m^2n^2p^2 = 2(x^2y^2z^2x*yx*zy*z) then x^2y^2z^2x*yx*zy*z = 1/2((xy)^2(yz)^2(xz)^2)
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Algebra Factor x^2y^2 x2 − y2 x 2 y 2 Since both terms are perfect squares, factor using the difference of squares formula, a2 −b2 = (ab)(a−b) a 2 b 2 = ( a b) ( a b) where a = x a = x and b = y b = y (xy)(x−y) ( x y) ( x y)X²y² = (x²2xyy²) 2xy we add (2xy) & (2xy) so that the both side will be equal And also to make a perfect square equation in the first equation = (xy)² 2xy we factor out the equation (x^22xyy^2) and then bring down (2xy) = (xy)² (√2xy)² we just bring down (xy)² and find a solution to find a square for 2xy to fit in the formula in factoring (ab)² (ab)² so we now
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