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

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

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

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

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

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

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

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

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

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

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

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

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1Yup! unless... using complex number...

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1yeah ... @RolyPoly

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome @tok1997
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