anonymous
  • anonymous
Divide Anyone want to help me out? x^2 -4/x-8/x-2 A.(x-2)(x+2)/x -8 B.x-8/x+2 <-- This is wrong C.x-2/x-8 D.x+2/x-8 I think it is C i'm just being for sure I failed math the first time trying not to do that again. XD
Mathematics
schrodinger
  • schrodinger
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Nnesha
  • Nnesha
is it \[\huge\rm \frac{ \frac{ x^2 -4}{ x-8 } }{ x-2 }\]
anonymous
  • anonymous
Yes it is.
Nnesha
  • Nnesha
alright show your work plz

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Nnesha
  • Nnesha
or explain how did you get C ?
anonymous
  • anonymous
To be honest guess it XD
Nnesha
  • Nnesha
i knew it don't guess and btw it's too hard to pick the option whn there are 4!
anonymous
  • anonymous
I Know but if you don't know how to solve it it's the next option.
Nnesha
  • Nnesha
\[\huge\rm \frac{ \frac{ x^2 -4}{ x-8 } }{ x-2 }\] first of all change division the multiplication so multiply first fraction by the reciprocal of the bottom fraction example \[\large\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d }}\] \[\bf \frac{ a }{ b } \times \frac{ d }{ c }\]
anonymous
  • anonymous
4|dw:1437412031089:dw| Like this
Nnesha
  • Nnesha
what's that ?
anonymous
  • anonymous
You said multiply by the reciprocal right.
Nnesha
  • Nnesha
yes reciprocal of the denominator
anonymous
  • anonymous
So the -2 and the -8?
Nnesha
  • Nnesha
and btw you can't separate x and -8
Nnesha
  • Nnesha
nope
Nnesha
  • Nnesha
(x^2-4) don't separate
anonymous
  • anonymous
Okay
anonymous
  • anonymous
so x^2-4*x^2-4?
Nnesha
  • Nnesha
\[\huge\rm \frac{ \frac{ x^2 -4}{ x-8 } }{\frac{ x-2}{1} }\] there is invisible one at the denominator of x-2
Nnesha
  • Nnesha
now multiply x^2-4/x-8 by the reciprocal of x-2/1
anonymous
  • anonymous
I see why i guessed it :/ I don't understand this..
Nnesha
  • Nnesha
what is reciprocal of x-2 /1 ?

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