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anonymous

  • 5 years ago

Partial fractions help.

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  1. amistre64
    • 5 years ago
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    .......yes??

  2. anonymous
    • 5 years ago
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    will someone please set up (4 5 or 6) and do one of 7 and 8 please?

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  3. amistre64
    • 5 years ago
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    lets do 4 :)

  4. amistre64
    • 5 years ago
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    3 ----------- (2x+1)(3x-2)

  5. amistre64
    • 5 years ago
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    We need on "fraction" for each factor in the bottom. A B 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2) is thsi correct?

  6. anonymous
    • 5 years ago
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    That looks good

  7. amistre64
    • 5 years ago
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    After we split it up into its "baser" units, we can try to "re-solve" it.

  8. amistre64
    • 5 years ago
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    A(3x-2) B(2x+1) ----------- + ----------- correct? (2x+1)(3x-2) (3x-2)(2x+1)

  9. amistre64
    • 5 years ago
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    the bottoms become redundant and we want the tops to equal out.

  10. anonymous
    • 5 years ago
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    Are you asking me whether you're right or are you just putting the questions there?

  11. anonymous
    • 5 years ago
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    guys no 4 is just setting up

  12. amistre64
    • 5 years ago
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    A(3x-2) + B(2x+1) = 3 lol.... they are really rhetorical, but I do tend to mess up alot ;)

  13. amistre64
    • 5 years ago
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    3x(A) -2(A) +2x(B) +1(B) = 3 combine like terms...

  14. amistre64
    • 5 years ago
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    3x(A) + 2x(B) -2(A) +1(B) ; factor out the variables now

  15. amistre64
    • 5 years ago
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    x(3A +2B) -1(2A +B) = 0x + 3

  16. amistre64
    • 5 years ago
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    3A +2B = 0 -2A -B = 3

  17. amistre64
    • 5 years ago
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    we can solve for A and B thru a system of equations now

  18. amistre64
    • 5 years ago
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    B = -3-2A 3A +2(-3-2A) = 0 3A -6 -4A = 0 -A = +6 A = -6

  19. amistre64
    • 5 years ago
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    B = -3-2(-6) B = -3+12 B = 9

  20. amistre64
    • 5 years ago
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    A B 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2) -3 9 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2)

  21. amistre64
    • 5 years ago
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    and thats our decomposition; any questions?

  22. anonymous
    • 5 years ago
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    for the set up only where should i stop?

  23. amistre64
    • 5 years ago
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    you should stop when all the possible "denominators" are accounted for: for example, if the denominator is (x+2)^3 we need to account for.. (x+2) and (x+2)^2 and (x+2)^3 as possible denominators in our setup

  24. anonymous
    • 5 years ago
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    what do we do for 3 terms in numerator?

  25. amistre64
    • 5 years ago
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    the 3 in the numberator is what we use to set up the condition for our "unknowns". A + B have to end up equaling 3 in the end.... its our goal, our guide, our calibration.

  26. amistre64
    • 5 years ago
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    If we get stuff like x^2(A+B) and there is no value for x^2 in our numberator, we set it up by 0x^2

  27. amistre64
    • 5 years ago
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    I might be missing a "+C" in there but I cant remember :)

  28. anonymous
    • 5 years ago
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    can you do 5 or 6 please? i am confused

  29. amistre64
    • 5 years ago
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    I can...... :)

  30. amistre64
    • 5 years ago
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    3x^2 +x -1 -------------- (x^2+2) (x^2+1) can we factor the bottom any more? I dont see that we can....

  31. anonymous
    • 5 years ago
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    no

  32. amistre64
    • 5 years ago
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    A B 3x^2 +x -1 ------- + ------- = ---------- (x^2+2) (x^2+1) yadayada Try to "solve" by getting like denominators now :) like normal...

  33. amistre64
    • 5 years ago
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    A(x^2+1) +B(x^2+2) = 3x^2 +x -1 ------------------ ----------- yadayada yadayada makes sense?

  34. anonymous
    • 5 years ago
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    yeah. thank you. I just needed the concept.

  35. amistre64
    • 5 years ago
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    now we can "turn" that left side into the right side like this :)

  36. amistre64
    • 5 years ago
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    x^2(A)+1(A) +x^2(B)+2(B) = 3x^2 +x -1 x^2(A) + x^2(B) +(A+2B) = 3x^2 +x -1 x^2(A+B) + (A+2B) = 3x^2 +x -1 ------------------ x^2(A+B) = 3x^2 A + B = 3 -------------------

  37. amistre64
    • 5 years ago
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    A + 2B = -1 A = -1 -2B ------------------

  38. amistre64
    • 5 years ago
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    -1 -2B +B = 3 -B = 4 B = -4

  39. anonymous
    • 5 years ago
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    i get it now.

  40. anonymous
    • 5 years ago
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    thank you very much

  41. anonymous
    • 5 years ago
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    gtg now

  42. anonymous
    • 5 years ago
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    seeya

  43. amistre64
    • 5 years ago
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    youre welcome :) bye..

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