Question

(2x-y)(2x+y)-(2x-y)^2

the answer is 4xy-2xy^2. I just don't know how they got that

Answer 1

both terms have a common factor of \(2x-7\) so you can "factor it out" to get
\[(2x-y)(2x+y-(2x-y))\]

Answer 2

on nvm it doesn't say factor, sorry
multiply out and combine like terms

Answer 3

\((2x-y)(2x+y)=4x^2-y^2\) and \((2x-y)^2=4x^2-4xy+y^2\)

Answer 4

use parentheses and compute
\[4x^2-y^2-(4x^2-4x+y^2)\]

Answer 5

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
\[4xy-2y^2\]

Answer 6

sorry, but I don understand I need to see it worked out step by step

Answer 7

i pretty much wrote every step out
except for the multiplications
we can do that if you like

Answer 8

yes please

Answer 9

\[(2x-y)(2x+y)-(2x-y)^2=(2x-y)(2x+y)-(2x-y)(2x-y)\] is the first step

Answer 10

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 \(-y\times y=-y^2\)

Answer 11

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
\[(2x+y)(2x-y)=4x^2-y^2\]

Answer 12

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
\[(2x-y)(2x-y)=4x^2-2xy-2xy+y^2=4x^2-4xy+y^2\]

Answer 13

you lost me at step 2

Answer 14

and the final step is to put it together and get
\[4x^2-y^2-(4x^2-4xy+y^2)\]

Answer 15

k is this \((2x+y)(2x-y)\) step 2?

Answer 16

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

Answer 17

mistake is here
\[(2x-y)(2x+y) - (2x-y)(2x-y) \\(4x^2+2xy-2xy-y^2)(-2x+y)(2x-y)\]

Answer 18

it should be
\[(2x-y)(2x+y) - (2x-y)(2x-y)\\ (4x^2+2xy-2xy-y^2)-\left((2x+y)(2x-y)\right)\]

Answer 19

thanks I get now