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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

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  1. 1on1
    • 3 years ago
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    Factor and simplify are different things.

  2. EpikRebel
    • 3 years ago
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    (a-b)^4n

  3. RolyPoly
    • 3 years ago
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    \[a^{4n}-b^{4n}\]\[=(a^{2n})^{2}-(b^{2n})^{2}=...\]

  4. EpikRebel
    • 3 years ago
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    If they share something, it can be factored.

  5. RolyPoly
    • 3 years ago
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    I'm sure if you expand (a-b)^4n, you won't get a^4n - b^4n

  6. tok1997
    • 3 years ago
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    okay yeah that was a stupid question :P I understand it now! thanks anyways guys!

  7. SheldonEinstein
    • 3 years ago
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    \[\large{\color{blue}{a^{4n}-b^{4n} = (a^{2n})^2 - (b^{2n})^2 = (a^{2n} + b^{2n} )(a^{2n}-b^{2n})}}\]

  8. RolyPoly
    • 3 years ago
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    a^2n - b^2n can be further factorised!

  9. SheldonEinstein
    • 3 years ago
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    The identity used here is \(\large{\color{red}{a^2-b^2=(a+b)(a-b)}}\)

  10. SheldonEinstein
    • 3 years ago
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    \[\large{(a^{2n}+b^{2n})(a^{2n}-b^{2n}) = (a^{2n}+b^{2n})[(a^n)^2 - (b^n)^2] \\= (a^{2n}+b^{2n}) (a^n + b^n)(a^n-b^n)}\]

  11. tok1997
    • 3 years ago
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    but a^(2n)+b^(2n) cant be furthur factorized?

  12. SheldonEinstein
    • 3 years ago
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    Yes @tok1997 a^(2n)+b^(2n) can not be further factorised but a^(2n)-b^(2n) can be.

  13. RolyPoly
    • 3 years ago
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    Yup! unless... using complex number...

  14. SheldonEinstein
    • 3 years ago
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    yeah ... @RolyPoly

  15. tok1997
    • 3 years ago
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    thanks guys :)

  16. SheldonEinstein
    • 3 years ago
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    You're welcome @tok1997

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