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anonymous

  • 5 years ago

Can (x+a)^2 be broken down into (x-a)(x+a) ?

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  1. anonymous
    • 5 years ago
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    no

  2. anonymous
    • 5 years ago
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    Not unless a = 0 \[(x+a)^2 = (x-a)(x+a) \rightarrow \frac{(x+a)^2}{(x+a)} = (x-a) \] \[\rightarrow x+a = x-a \rightarrow a = -a\]

  3. anonymous
    • 5 years ago
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    Okydokey.

  4. anonymous
    • 5 years ago
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    a=0 is the trivial answer. For all x not equal to zero, \[(x+a)^{2}=(x+a)(x+a)\]

  5. anonymous
    • 5 years ago
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    Sorry, made a typo! I meant for all "a" not equal to zero.

  6. anonymous
    • 5 years ago
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    Ok, I don't know if you're a big fan of integration but I'm sure you'll be able to answer this. If I have the integral of dx / (X^4 - a^4), does that break down into dx / (x-a)^2(x+a)^2 . And then after that using what you've told me, does it break down further into dx / (x-a)(x+a)(x+a)(x+a) ?

  7. anonymous
    • 5 years ago
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    And we're not told what a is equal to.

  8. anonymous
    • 5 years ago
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    let \[u = x ^{2}\] then let \[b = a ^{2}\] now you have \[u^{2}-a ^{2}\] factor out -1 now you have \[-(a ^{2}-u ^{2})\] the integral becomes \[-1\int\limits du/(a ^{2} - u ^{2})\] Now you can look up the integral in a table of integrals

  9. anonymous
    • 5 years ago
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    made a typo\[-(b ^{2}-u ^{2)}\] now the integral becomes \[-1 \int\limits du / (b ^{2} - u ^{2} )\] This integral is listed in any table of integrals.

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