rannsan
completely simplify (x2)/(8x^2 32)



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Outkast3r09
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\[\frac{x2}{8x^232}\]

apoorwa
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(x2)/(8(x^24))

apoorwa
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(x2)/(8(x+2)(x2))
\[\frac{ (x2) }{ (8(x+2)(x2)) }\]

Outkast3r09
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first factor out an 8 out of the bottom
\[\frac{x2}{8(x^24)}\]
next use \[(ab)^2 =(ab)(a+b)\]
when a and b are perfect squares

apoorwa
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cancel the common now

Outkast3r09
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\[a^2b^2=(a+b)(ab)\]

rannsan
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ok

rannsan
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1/8(x+2) is this right?

rannsan
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= (x2)/8(x^24)
= (x2)/8(x+2)(x2)
= 1/8(x+2)

apoorwa
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yes thats absolutely corret:)

rannsan
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Thank you