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how can i factor:
a^(4n)b^(4n)
Seems to me its already in its simplest form
 one year ago
 one year ago
how can i factor: a^(4n)b^(4n) Seems to me its already in its simplest form
 one year ago
 one year ago

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1on1Best ResponseYou've already chosen the best response.0
Factor and simplify are different things.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.1
\[a^{4n}b^{4n}\]\[=(a^{2n})^{2}(b^{2n})^{2}=...\]
 one year ago

EpikRebelBest ResponseYou've already chosen the best response.0
If they share something, it can be factored.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.1
I'm sure if you expand (ab)^4n, you won't get a^4n  b^4n
 one year ago

tok1997Best ResponseYou've already chosen the best response.0
okay yeah that was a stupid question :P I understand it now! thanks anyways guys!
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
\[\large{\color{blue}{a^{4n}b^{4n} = (a^{2n})^2  (b^{2n})^2 = (a^{2n} + b^{2n} )(a^{2n}b^{2n})}}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.1
a^2n  b^2n can be further factorised!
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
The identity used here is \(\large{\color{red}{a^2b^2=(a+b)(ab)}}\)
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
\[\large{(a^{2n}+b^{2n})(a^{2n}b^{2n}) = (a^{2n}+b^{2n})[(a^n)^2  (b^n)^2] \\= (a^{2n}+b^{2n}) (a^n + b^n)(a^nb^n)}\]
 one year ago

tok1997Best ResponseYou've already chosen the best response.0
but a^(2n)+b^(2n) cant be furthur factorized?
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
Yes @tok1997 a^(2n)+b^(2n) can not be further factorised but a^(2n)b^(2n) can be.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.1
Yup! unless... using complex number...
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
yeah ... @RolyPoly
 one year ago

SheldonEinsteinBest ResponseYou've already chosen the best response.1
You're welcome @tok1997
 one year ago
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