Question

(3xy-5zx) * (2zx-8xy)​

Answer 1

GivEn:

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• (3xy-5zx)(2zx-8xy)

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To find:

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• The product of (3xy-5zx) and (2zx-8xy).

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Solution:

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~ Here first we we'll multiply the first term of first factor with both the terms of second factor. Then, Multiply the second term of first factor with both the terms of second factor.

Here,

• First factor = (3xy-5zx)
• Second factor = (2zx-8xy)
• First terms of both factors = 3xy,2zx
• Second terms of both factors = 5zx,8xy

☯ Now, Let's find the product!

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$\implies \sf (3xy-5zx)(2zx-8xy) \\ \\ \\ \implies \sf 6 {x}^{2} yz - 24 {x }^{2} {y}^{2} - 10 {x}^{2} {z}^{2} + 40 {x}^{2} yz \\ \\ \\ \implies \sf - 24 {x }^{2} {y}^{2} - 10 {x}^{2} {z}^{2} + 40 {x}^{2} yz + 6 {x}^{2} yz \\ \\ \\\implies \pmb{ \sf \purple{- 24 {x }^{2} {y}^{2} - 10 {x}^{2} {z}^{2} + 46 {x}^{2} yz }}$

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$\therefore\:{\underline{\sf{Thus,\:the\: product\:of\:(3xy-5zx)\: and \: (2zx-8xy)\:\:is\: {\pmb{\sf{- 24 {x }^{2} {y}^{2} - 10 {x}^{2} {z}^{2} + 46 {x}^{2} yz }}}.}}}$

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$\boxed{\begin{array}{cc}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}$