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Partial Fractions - irreducibility question

  1. Oct 16, 2014 #1

    RJLiberator

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    1. x^2-x+1

    Is this factorable?

    My initial thinking is NO. However, I can complete the square and it becomes (x-1/2)^2-3/4, but this doesn't seem to help me. Would this be considered factorable?

    2. Turn 1/x^2-x+1 into partial fractions

    Clearly, after I answer #1 correctly, #2 will be more clear. But under my initial assumption that #1 is NOT reducible, then how would I turn 1/x^2-x+1 into a partial fraction?
    I am getting 1 = Ax+b where b = 0 and A = 1/x but that doesn't seem at all correct.

    Thanks for any guidance / help.
     
  2. jcsd
  3. Oct 16, 2014 #2

    Mark44

    Staff: Mentor

    You have an error in your work. You should have (x - 1/2)2 + 3/4, not (x - 1/2)2 - 3/4. This quadratic is not factorable into factors with real number coefficients.
    This expression needs parentheses around the entire denominator. What you wrote is this:
    $$\frac{1}{x^2} - x + 1$$
    No, it's not. The coefficients A and B should be constants, not expressions that involve a variable.

    BTW, what happened to the homework template? You should not delete its parts.
     
  4. Oct 16, 2014 #3

    RJLiberator

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    Fair enough.

    Do I use the denominator of x^2-x+1 or (x-1/2)^2 +3/4 to use partial fractions?
     
  5. Oct 16, 2014 #4

    LCKurtz

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    You don't turn ##\frac 1 {x^2-x+1}## into a partial fraction. In the second form above, where you have completed the square, it is already ready to integrate using an arctangent.
     
  6. Oct 16, 2014 #5

    RJLiberator

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    LcKurtz, thank you my brother for the help verifying my thoughts.
     
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