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Math Help - Expressing in partial fractions

  1. #1
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    Smile Expressing in partial fractions

    (9x^2) / [(x -1)^2 * (x + 2)]

    It's part of a set of partial fractions I have to do.. I can't seem to solve this?
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    Quote Originally Posted by erika View Post
    (9x^2) / [(x -1)^2 * (x + 2)]

    It's part of a set of partial fractions I have to do.. I can't seem to solve this?
    \frac {9x^2}{(x - 1)^2 (x + 2)} = \frac {A}{x - 1} + \frac {B}{(x - 1)^2} + \frac {C}{x + 2}

    Can you take it from here?
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  3. #3
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    Meh I did that and then tried substituting values for x but it didn't seem to work
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by erika View Post
    Meh I did that and then tried substituting values for x but it didn't seem to work
    Substitute values of x? do you know the method of partial fractions?

    multiply through by the denominator of the left side, we get:


    9x^2 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2

    Can you take it from here? You can multiply out and equate coefficients, or you can choose convenient values of x that wipes out two variables at a time
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  5. #5
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    sub x: 1
    9 = 3B
    3 = B

    sub x: -1
    36 = 9C
    4 = C

    By comparing coefficient of x^2:
    9 = A + C
    A = 5

    Is that correct?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by erika View Post


    sub x: 1
    9 = 3B
    3 = B

    sub x: -1
    36 = 9C
    4 = C

    By comparing coefficient of x^2:
    9 = A + C
    A = 5

    Is that correct?
    Yes! Good Job!
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  7. #7
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    Thank you!
    One last question...

     <br />
\frac {17x^2+23x+12}{(3x+4)(x^2+4)} = \frac{A}{3x+4} + \frac{B}{x^2+4}<br />

    By cover-up rule, I got
     <br />
A = \frac {17(-4/3)^2+23(-4/3)+12}{(-4/3)^2+4}<br />
= 2<br />

    But since  (x^2 +4) isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
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  8. #8
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    Quote Originally Posted by erika View Post
    Thank you!
    One last question...

     <br />
\frac {17x^2+23x+12}{(3x+4)(x^2+4)} = \frac{A}{3x+4} + \frac{B}{x^2+4}<br />

    By cover-up rule, I got
     <br />
A = \frac {17(-4/3)^2+23(-4/3)+12}{(-4/3)^2+4}<br />
= 2<br />
    what is this "cover-up" rule you are referring to?



    But since  (x^2 +4) isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
    yes, but you have to change the numerator. The numerator must always be one degree less than the denominator. so for the part where the denominator is x^2 + 4 you must write:

    \frac {Bx + C}{x^2 + 4}
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