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sat_chen

  • 3 years ago

Partial integration question

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  1. sat_chen
    • 3 years ago
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    hold on i didnt know you couldnt post pic from iphone jsut 1 sec

  2. hartnn
    • 3 years ago
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    draw your question, maybe...

  3. sat_chen
    • 3 years ago
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    i have done a lot of the steps so just 1 sec

  4. sat_chen
    • 3 years ago
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    but heres the question |dw:1359962006954:dw|

  5. hartnn
    • 3 years ago
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    that will be pretty exhausting..

  6. hartnn
    • 3 years ago
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    x^4 (x+3) A/x+B/x^2+C/x^3+D/x^4 +E/(x+3)

  7. sat_chen
    • 3 years ago
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  8. sat_chen
    • 3 years ago
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    do you see my steps ill try posting another pic

  9. hartnn
    • 3 years ago
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    yes, D=1, E=4 are correct.

  10. sat_chen
    • 3 years ago
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    yea i got that far phew but how do i solve for the A B and C after that?

  11. hartnn
    • 3 years ago
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    you'll need to put 3 different values of x say x=1,-1,2 then you get 3 equations in A,B, C with 3 unknowns...

  12. sat_chen
    • 3 years ago
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    can you show me what you mean i dont get what you mean by that

  13. sat_chen
    • 3 years ago
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    should really have listened in my algebra classes sorry :(

  14. hartnn
    • 3 years ago
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    \(Ax^3(x+3)+Bx^2(x+3)+Cx^3(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3\) put x = 1, to get 1 equation in A,B, C

  15. sat_chen
    • 3 years ago
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    it should be Cx (x+3)

  16. hartnn
    • 3 years ago
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    another method is comparing the co-efficients to get 3 equations... yes, typo...it should be Cx(x+3)

  17. hartnn
    • 3 years ago
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    like co-efficient of x^4 will be A+4 and no right side, its 9 so, A+4= 9 gives A=5

  18. hartnn
    • 3 years ago
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    i think you should proceed with comparing the co-efficients..

  19. sat_chen
    • 3 years ago
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    so if you plug in x = 1 you get 4A + 4B+ 4C = 16

  20. hartnn
    • 3 years ago
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    i think comapring the co-efficients will give you answer much faster, if you have understood it.

  21. sat_chen
    • 3 years ago
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    sorry internet disconnected agian

  22. hartnn
    • 3 years ago
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    co-efficient of x^3 = 3A+B = 15+B on left on right its 17 so, 15+B = 17 gives B=3

  23. hartnn
    • 3 years ago
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    forget about substitution...

  24. sat_chen
    • 3 years ago
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    ok so what you are saying is that for A you A + 4 = 9 then for B would you B + 3 =17 then for C would you C + 2 =3

  25. sat_chen
    • 3 years ago
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    where did you get the|dw:1359963485018:dw|

  26. hartnn
    • 3 years ago
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    it should be B+3A =17

  27. sat_chen
    • 3 years ago
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    then i also guess i dont get how you got the A + 4 either because my logic was wrong

  28. hartnn
    • 3 years ago
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    |dw:1359963578884:dw|

  29. sat_chen
    • 3 years ago
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    not getting the |dw:1359963731639:dw|

  30. hartnn
    • 3 years ago
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    \(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3\)

  31. sat_chen
    • 3 years ago
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    do you have a formula or something that tells us that or how does that work?

  32. hartnn
    • 3 years ago
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    \(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\ (A+4)x^4+(3A+B)x^3+....=...\)

  33. hartnn
    • 3 years ago
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    comapring the c-efficiens on left and right, A+4=9 3A+B=17 did you get this ?

  34. sat_chen
    • 3 years ago
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    its not showing it in nice formatting so im writing it down just 1 sec

  35. hartnn
    • 3 years ago
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    Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4=9x^4+17x^3+3x^2−8x+3 Ax^4+3Ax^3+Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4=9x^4+17x^3+3x^2−8x+3 (A+4)x^4+(3A+B)x^3+....=9x^4+17x^3+...

  36. hartnn
    • 3 years ago
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    i just combined like terms on left..

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