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Math Help - Partial Fraction Decomposition

  1. #1
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    Partial Fraction Decomposition

    (8x+12)/x(x+4) =A/x+B/(x+4)

    Looking for some help.

    This is how I'm approaching it so far and not sure I'm doing it correctly.

    8x+12=A(x+4)+B(x)

    8x+12=(A+B)x+(4A+B)
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  2. #2
    Junior Member slider142's Avatar
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    You are proceeding correctly. Your next step would be to note that the coefficients of powers of x must be equivalent on each side of the equation, as A and B are numbers, not functions of x, and cannot introduce an x term.
    That is, you are now to solve the two simultaneous equations:
    8 = A + B
    12 = 4A + B
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  3. #3
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    So then x=1, A = 4/3, B = 20/3 ??
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  4. #4
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    Quote Originally Posted by RBlax View Post
    (8x+12)/x(x+4) =A/x+B/(x+4)
    Is the left side correctly stated: (8x+12)/x(x+4) means [(8x+12)/x] * (x+4)
    or do you mean (8x+12) / [x(x+4)] ?
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  5. #5
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    (8x+12) over (x(x+4)) =)
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  6. #6
    Junior Member slider142's Avatar
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    My apologies; I did not actually check your algebra in your first post.
    In the line:
    8x + 12 = A(x + 4) + Bx
    the next line should be:
    8x + 12 = (A + B)x + 4A
    so that the simultaneous equations are:
    A + B = 8
    4A = 12.
    Note that you need not rely on us to check whether your solution is correct; simply add the resulting fractions together replacing A and B as necessary and see if you get the correct result.
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  7. #7
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    Quote Originally Posted by RBlax View Post
    (8x+12)/x(x+4) =A/x+B/(x+4)

    Looking for some help.

    This is how I'm approaching it so far and not sure I'm doing it correctly.

    8x+12=A(x+4)+B(x)

    8x+12=(A+B)x+(4A+B)
    As you have been told, two polynomials are the same for all x if and only if they have the same coefficients: you must have A+ B= 8 and 4A+ B= 12.

    Here's a slightly different way: going back to 8x+ 12= A(x+ 4)+ Bx, since this is true for all x, you can choose whatever values you like for x. In particular, taking x= -4 and then x= 0 gives you two very simple equations!

    (You can't always find values of x that will make it that easy- in that case, you set coefficients equal or just choose two simple values for x to get two equations to solve for A and B.)
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