anonymous
  • anonymous
Write the partial fraction decomposition of the rational expression.
Mathematics
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anonymous
  • anonymous
Write the partial fraction decomposition of the rational expression.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/5224e4dbe4b0750826e245ae-gponce28-1378149870417-f1q1g1.jpg
anonymous
  • anonymous
hartnn
  • hartnn
know how to start?? \(\large \dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-4}\)

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anonymous
  • anonymous
what is that
hartnn
  • hartnn
partial fraction decomposition of your expression. once you find A,B,C you're done
hartnn
  • hartnn
\(\large \dfrac{18x^2 -68x+24}{x(x-2)(x-4)} = \dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-4}\)
hartnn
  • hartnn
now simplify the right side, so that it has one common denominator
hartnn
  • hartnn
like 1/a + 1/b = (b+a)/ab
hartnn
  • hartnn
let me give you a start denominator = x(x-2)(x-4) numerator = A (x-2)(x-4) + B.................
anonymous
  • anonymous
im so cunfused
hartnn
  • hartnn
have you solved a similar problem before?
anonymous
  • anonymous
no i just srted this class
hartnn
  • hartnn
too early to take that example then
hartnn
  • hartnn
start with something easier \(\dfrac{2x+3}{(x+1)(x+2)} \)
anonymous
  • anonymous
ooh;
hartnn
  • hartnn
to decompose that into partial fractions means, we need to have fractions with linear denominators \(\dfrac{2x+3}{(x+1)(x+2)} = \dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}\)
hartnn
  • hartnn
then find A,B with any method that you prefer. I tried to discuss One of the method earlier
hartnn
  • hartnn
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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