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anonymous

  • one year ago

Write the partial fraction decomposition of the rational expression.

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  1. anonymous
    • one year ago
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    @Nnesha

  2. hartnn
    • one year ago
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    know how to start?? \(\large \dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-4}\)

  3. anonymous
    • one year ago
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    what is that

  4. hartnn
    • one year ago
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    partial fraction decomposition of your expression. once you find A,B,C you're done

  5. hartnn
    • one year ago
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    \(\large \dfrac{18x^2 -68x+24}{x(x-2)(x-4)} = \dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-4}\)

  6. hartnn
    • one year ago
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    now simplify the right side, so that it has one common denominator

  7. hartnn
    • one year ago
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    like 1/a + 1/b = (b+a)/ab

  8. hartnn
    • one year ago
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    let me give you a start denominator = x(x-2)(x-4) numerator = A (x-2)(x-4) + B.................

  9. anonymous
    • one year ago
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    im so cunfused

  10. hartnn
    • one year ago
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    have you solved a similar problem before?

  11. anonymous
    • one year ago
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    no i just srted this class

  12. hartnn
    • one year ago
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    too early to take that example then

  13. hartnn
    • one year ago
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    start with something easier \(\dfrac{2x+3}{(x+1)(x+2)} \)

  14. anonymous
    • one year ago
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    ooh;

  15. hartnn
    • one year ago
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    to decompose that into partial fractions means, we need to have fractions with linear denominators \(\dfrac{2x+3}{(x+1)(x+2)} = \dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}\)

  16. hartnn
    • one year ago
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    then find A,B with any method that you prefer. I tried to discuss One of the method earlier

  17. hartnn
    • one year ago
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    taking a common denominator on right side \(\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)} = \dfrac{A(x+1)+B(x+2)}{(x+1)(x+2)}\) makes sense?

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