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tok1997
Group Title
how can i factor:
a^(4n)b^(4n)
Seems to me its already in its simplest form
 2 years ago
 2 years ago
tok1997 Group Title
how can i factor: a^(4n)b^(4n) Seems to me its already in its simplest form
 2 years ago
 2 years ago

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1on1 Group TitleBest ResponseYou've already chosen the best response.0
Factor and simplify are different things.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
\[a^{4n}b^{4n}\]\[=(a^{2n})^{2}(b^{2n})^{2}=...\]
 2 years ago

EpikRebel Group TitleBest ResponseYou've already chosen the best response.0
If they share something, it can be factored.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I'm sure if you expand (ab)^4n, you won't get a^4n  b^4n
 2 years ago

tok1997 Group TitleBest ResponseYou've already chosen the best response.0
okay yeah that was a stupid question :P I understand it now! thanks anyways guys!
 2 years ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
a^2n  b^2n can be further factorised!
 2 years ago

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

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
\[\large{(a^{2n}+b^{2n})(a^{2n}b^{2n}) = (a^{2n}+b^{2n})[(a^n)^2  (b^n)^2] \\= (a^{2n}+b^{2n}) (a^n + b^n)(a^nb^n)}\]
 2 years ago

tok1997 Group TitleBest ResponseYou've already chosen the best response.0
but a^(2n)+b^(2n) cant be furthur factorized?
 2 years ago

SheldonEinstein Group TitleBest 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.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Yup! unless... using complex number...
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
yeah ... @RolyPoly
 2 years ago

tok1997 Group TitleBest ResponseYou've already chosen the best response.0
thanks guys :)
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
You're welcome @tok1997
 2 years ago
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