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Math Help - help with partial fractions

  1. #16
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    Ok, so do it. Type up the next step.
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  2. #17
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    There is no number that they go into from what I can see
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  3. #18
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    Any integer can go into something! If you mean that no number goes into any of them, I would agree with you. That's why I picked primed numbers. However, you don't need to do that; instead, you need to find the least common multiple of all the denominators. How do you do that?
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  4. #19
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    not sure, its been about 15 year since I did basic maths. I think the best thing would be to start backwards. I know how to work out the partial fractions for 2 and 3 parts. But not sure how they were derived. for example I know from memory that A/(S-1) + B/(S-1)^2 + C(S+3) becomes A(S-1)(S+3) +B(S+3) + C(S-1)^2. I dont know why, I just memorised it.
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  5. #20
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    If you'll just let me, I'll teach you. But if you refuse to attempt what I suggest, then there's not much I can do for you.

    What is the least common multiple of 2 and 3?
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  6. #21
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  7. #22
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    Good. What is the LCM of 3, 7, and 9?
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  8. #23
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  9. #24
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    Right. Now, how do you add

    \dfrac{1}{2}+\dfrac{1}{3}?
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  10. #25
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    5/6
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  11. #26
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    Right. So you added this way, correct?

    \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{3}{6}+\dfrac{2}{6  }=\dfrac{3+2}{6}=\dfrac{5}{6}?

    Now, add these fractions together:

    \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}.

    What do you get? Show all your steps just like I did above with the 1/2 + 1/3.
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  12. #27
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    1/2 + 1/3 +1/5 = 15/30 + 10/30 + 6/30 = 15 + 10 + 6/30 = 31/30
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  13. #28
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    Good. Now, add these up:

    \dfrac{1}{2x^{2}+x+5}+\dfrac{2}{x+1}.
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  14. #29
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    this is where I dont know what to do. I know u factorize the donominator on the left to put it in 2 parts then join them to the right donominator to make 3 but not sure about the numerator.
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  15. #30
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    In general, you'd be right. However, in this case, the left-hand fraction doesn't factor over the reals (the discriminant b^{2}-4ac<0.) Therefore, you have to view the entire denominator as a factor. In other words, the entire denominator of the left-hand fraction acts like one of the prime numbers in the denominators of the fractions in post # 8. How does that change things for you?
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