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Math Help - A little minor question on Partial Fraction

  1. #1
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    A little minor question on Partial Fraction

    Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.

    After that I found A=5 and have unknown B. What should I sub x now?
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    Quote Originally Posted by Punch View Post
    Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.

    After that I found A=5 and have unknown B. What should I sub x now?
    It would help if you post the entire question...
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    Erm, not asking a question but instead clarifying something. Okay probably if i explain it this way it would be easier to understand.

    What I am asking is, what should I substitute x as when I run out of values to substitute x as?
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    Quote Originally Posted by Punch View Post
    Erm, not asking a question but instead clarifying something. Okay probably if i explain it this way it would be easier to understand.

    What I am asking is, what should I substitute x as when I run out of values to substitute x as?
    The value of x that you substitute is different for every problem.

    This is why I need to see the problem.
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  5. #5
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    Oh, so this is the case...
    Alright then this poses a new problem and a new question in my head, how do I know what to substitute x as when I run out of values to substitute x as?
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    Quote Originally Posted by Punch View Post
    Oh, so this is the case...
    Alright then this poses a new problem and a new question in my head, how do I know what to substitute x as when I run out of values to substitute x as?
    All of these questions will be answered when you POST A PROBLEM so that I CAN DEMONSTRATE HOW TO SOLVE IT.
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    Express in partial fraction \frac{17x^2+23x+12}{(3x+4)(x^2+4)}
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    Quote Originally Posted by Punch View Post
    Express in partial fraction \frac{17x^2+23x+12}{(3x+4)(x^2+4)}
    Checklist:

    1. The numerator has a degree less than the denominator. (Tick)

    2. The denominator can be factorised. (Tick)


    Now, we require a partial fraction of the form

    \frac{A}{3x + 4} + \frac{Bx + C}{x^2 + 4} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}.

    The numerator is always of one less degree than the denominator.


    Getting the common denominator:

    \frac{A(x^2 + 4) + (Bx + C)(3x + 4)}{(3x + 4)(x^2 + 4)} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}.


    Can you see that since the denominators are equal, so must the numerators...

    So A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12.


    In this case, there does not exist an x that you can use to eliminate the A.

    So we will use another method. Expand all the brackets...

    Ax^2 + 4A + 3Bx^2 + 4Bx + 3Cx + 4C = 17x^2 + 23x + 12

    Equating like powers of x, we get

    (A + 3B)x^2 + (4B + 3C)x + 4A + 4C = 17x^2 + 23x + 12.


    Can you see that for this equation to be true, then

    A + 3B = 17

    4B + 3C = 23

    4A + 4C = 12.


    Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here?
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    Well, one of the other method is to substitute x= -4/3?
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    Quote Originally Posted by Punch View Post
    Well, one of the other method is to substitute x= -4/3?
    Yes, if you substitute -\frac{4}{3} you can eliminate B and C to find A.

    But there does not exist any number that will be able to eliminate A to find B and C.


    Using your method...

    A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12

    A\left[\left(-\frac{4}{3}\right)^2 + 4\right] + (Bx + C)\left[3\left(-\frac{4}{3}\right) + 4\right] = 17\left(-\frac{4}{3}\right)^2 + 23\left(-\frac{4}{3}\right) + 12

    \frac{52}{9}A = \frac{272}{9} - \frac{92}{3} + 12

    \frac{52}{9}A = \frac{272}{9} - \frac{276}{9} + \frac{108}{9}

    \frac{52}{9}A = \frac{104}{9}

    A = 2.


    Now that you have A you should be able to solve the simultaneous equations I gave you easier.

    But like I said, in this case, the ONLY method that will get you all three unknowns is to equate like powers of x.
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    However I have seen some examples in which they substitute x=0 to find other unknowns
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    You substitute whatever will make the coefficients of A, B, C, etc = 0.

    So in your case, we wanted the coefficient of Bx + C to become 0.

    So we let 3x + 4 = 0

    3x = -4

    x = -\frac{4}{3}.


    Notice that the coefficient of A is x^2 + 4. To try to eliminate A, we would need to let x^2 + 4 = 0

    x^2 = -4.

    Is it possible to square a number and end up with something negative?


    You would only substitute x = 0 if one of your coefficients happened to be x.
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    Substituting 0 means finding c and so I need to substitute 1 after 0?
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    You would not be able to find C by substituting x = 0 or x = 1. You will still have A in the equation.


    Edit: Actually, if you have already found A, then yes, you can substitute x = 0.


    You have

    A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12.

    Substituting A = 2 and x = 0 we find

    2(0^2 + 4) + [B(0) + C][3(0) + 4] = 17(0)^2 + 23(0) + 12

    8 + 4C = 12

    4C = 4

    C = 1.
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  15. #15
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    Look you have already found a to be 2 so the a part will make a value. which will help in deducing what c is... and after finding what c is, we substitute x=1 to find b.
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