anonymous
  • anonymous
(2x-y)(2x+y)-(2x-y)^2 the answer is 4xy-2xy^2. I just don't know how they got that
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
both terms have a common factor of \(2x-7\) so you can "factor it out" to get \[(2x-y)(2x+y-(2x-y))\]
anonymous
  • anonymous
on nvm it doesn't say factor, sorry multiply out and combine like terms
anonymous
  • anonymous
\((2x-y)(2x+y)=4x^2-y^2\) and \((2x-y)^2=4x^2-4xy+y^2\)

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anonymous
  • anonymous
use parentheses and compute \[4x^2-y^2-(4x^2-4x+y^2)\]
anonymous
  • anonymous
except don't make the typo i made, it is \[4x^2-y^2-(4x^2-4xy+y^2)\] distribute minus sign to get \[4x^2-y^2-4x^2+4xy-y^2\] and finally combine like terms \[4x^2-4x^2=0\] and \(y^2-y^2=-2y^2\) leavin g \[4xy-2y^2\]
anonymous
  • anonymous
sorry, but I don understand I need to see it worked out step by step
anonymous
  • anonymous
i pretty much wrote every step out except for the multiplications we can do that if you like
anonymous
  • anonymous
yes please
anonymous
  • anonymous
\[(2x-y)(2x+y)-(2x-y)^2=(2x-y)(2x+y)-(2x-y)(2x-y)\] is the first step
anonymous
  • anonymous
so now we have to multiply \[(2x+y)(2x-y)\] needs 4 multiplications sometimes called "first outer inner last" the "first" is \(2x\times 2x=4x^2\) the "outer" is \(-2xy\) the "inner" is \(2xy\) and the last is \(-y\times y=-y^2\)
anonymous
  • anonymous
putting it together gives \[(2x+y)(2x-y)=4x^2+2xy-2xy-y^2\] and the middle terms add up to zero, so you ge t \[(2x+y)(2x-y)=4x^2-y^2\]
anonymous
  • anonymous
now we have to repeat the process with \[(2x-y)(2x-y)\]but this time the middle terms will not add up to zero doing it a bit faster gives \[(2x-y)(2x-y)=4x^2-2xy-2xy+y^2=4x^2-4xy+y^2\]
anonymous
  • anonymous
you lost me at step 2
anonymous
  • anonymous
and the final step is to put it together and get \[4x^2-y^2-(4x^2-4xy+y^2)\]
anonymous
  • anonymous
k is this \((2x+y)(2x-y)\) step 2?
anonymous
  • anonymous
I don't understand I got this (2x-y)(2x+y) - (2x-y) ^2 (2x-y)(2x+y) - (2x-y)(2x-y) (4x^2+2xy-2xy-y^2)(-2x+y)(2x-y) (4x^2-y^2)-4x^2-2xy+2xy-Y^2 4x^2-y^2-4x^2-Y^2 I know this is wrong but dont understand what im suppose to do
anonymous
  • anonymous
mistake is here \[(2x-y)(2x+y) - (2x-y)(2x-y) \\(4x^2+2xy-2xy-y^2)(-2x+y)(2x-y)\]
anonymous
  • anonymous
it should be \[(2x-y)(2x+y) - (2x-y)(2x-y)\\ (4x^2+2xy-2xy-y^2)-\left((2x+y)(2x-y)\right)\]
anonymous
  • anonymous
thanks I get now

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