megannicole51
  • megannicole51
split into partial fractions....((5)/(y^3-25y))
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/partial-fraction-expansion/v/partial-fraction-expansion-1
zepdrix
  • zepdrix
\[\Large \frac{5}{y^3-25y} \quad=\quad \frac{5}{y(y^2-5^2)}\]This might be a good first step right? :) Looks like we can further factor though. Remember how to break down the `difference of squares` ?
megannicole51
  • megannicole51
(A/y)+(B/y-5)+(C/(y-5)^2)?

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zepdrix
  • zepdrix
Woops you may have factored your denominator incorrectly, let's see.
zepdrix
  • zepdrix
\[\Large \frac{5}{y(y^2-5^2)} \ne\frac{5}{y(y-5)^2}\]
megannicole51
  • megannicole51
thats how my professor broke down a problem in my notes....probably applies to a different type of problem
zepdrix
  • zepdrix
yah probably D:
zepdrix
  • zepdrix
We use this rule for factoring the difference of squares: \[\large \color{royalblue}{a^2-b^2=(a-b)(a+b)}\] So what does that do to our\[\Large y^2-25^2\]
zepdrix
  • zepdrix
woops* \(\Large y^2-5^2\)
megannicole51
  • megannicole51
im confused....why are you doing that?
anonymous
  • anonymous
watch the vid if you dont get it by the first vid there are 2 more vids :)
zepdrix
  • zepdrix
We would like to be able to write the denominator as `linear factors` if we're able to. Which means all factors would contain y to the first power. We're starting with y^3 and trying to break it down to y^1's. It will make our partial fraction decomposition much easier.
megannicole51
  • megannicole51
@pgpilot326
Hero
  • Hero
The first step is to factor the denominator: \[y^3-25y = y(y^2 - 25) = y(y + 5)(y - 5)\] Afterwards, you write the partial expansion in the form: \[\frac{5}{y(y + 5)(y - 5)} = \frac{A}{y - 5} + \frac{B}{y} + \frac{C}{y + 5}\]
megannicole51
  • megannicole51
i put that earlier and someone said it was wrong?
megannicole51
  • megannicole51
actually my numerators were a bit different
anonymous
  • anonymous
look at the denominator and factor, right? \[ y^{3} -25y = y\left(y^{2}-25\right) = y(y-5)(y+5)\] now look over those papers and figure out what to do. if you get stuck, message me or reply addressing me. You've got this, girl!
Hero
  • Hero
Next multiply both sides by \(y(y + 5)(y - 5))\), then simplify: \[5 = y(y + 5)A + (y - 5)(y + 5)B + y(y - 5)C\]
megannicole51
  • megannicole51
why did you set the denominators up the way u did?
Hero
  • Hero
@megannicole51, because that's the form you want when expanding to partial fractions.
Hero
  • Hero
Now you have to expand the right side
Hero
  • Hero
The right side expands to: \(y(y + 5)A + (y + 5)(y - 5)B + y(y - 5)C\) \(= (y^2 + 5y)A + (y^2 - 25)B + (y^2 - 5y)C\) \(=Ay^2 + A5y + By^2 - 25B + Cy^2 - C5y\) \(=Ay^2 + By^2 + Cy^2 + A5y - C5y - 25B\) \(=(A + B + C)y^2 + (A - C)5y - 25B\)
Hero
  • Hero
\( = (A + B + C)y^2 + (5A - 5C)y - 25B\)
megannicole51
  • megannicole51
i have to go to class but ill be back to take a look at ur expansion!
Hero
  • Hero
Basically you end up with this: \(5 = (A + B + C)y^2 + (5A - 5C)y - 25B\) And then you set up a system like so: \(0 = A + B + C\) \(0 = 5A - 5C\) \(5 = -25B\)
Hero
  • Hero
Then after solving the system you end up with: \[A = \frac{1}{10}\]\[B = -\frac{1}{5}\]\[C = \frac{1}{10}\]
Hero
  • Hero
And the final form you get should be: \[\frac{5}{y(y + 5)(y - 5)} = \frac{1}{10(y - 5)} -\frac{1}{5y} + \frac{1}{10(y + 5)}\]
megannicole51
  • megannicole51
0=A+B+C 0=5A−5C 5=−25B how do you know this?
megannicole51
  • megannicole51
@Hero
anonymous
  • anonymous
3eq with 3 unknowns : start with 5=−25B divide both sides by -25 B = -1/5 now from the second you get A=C plug B=-1/5 and A=C into the first 0=A-1/5+A 2A = 1/5 A = 1/10 nnow since A=C C=1/10
anonymous
  • anonymous
oh that is not what you asked.
megannicole51
  • megannicole51
my question is just how do u know what those equal? 0=A+B+C 0=5A−5C 5=−25B
anonymous
  • anonymous
so you had: 5 = (A+B+C)y^2 + (5A-5C)y -25B
anonymous
  • anonymous
as you can see at the left side you have only a number (independent of y)
anonymous
  • anonymous
so both of the terms that multiply y and y^2 at the right side must be 0
anonymous
  • anonymous
A+B+C = 0 and 5A-5C = 0
anonymous
  • anonymous
now what is left is the number at the right side : -25B it must be equal to the number at the left : 5 -25B = 5
megannicole51
  • megannicole51
oooh okay
megannicole51
  • megannicole51
Thank you to everyone who helped!
Hero
  • Hero
Sorry, I stepped away from my computer for a moment. @megannicole51, do you get everything else?
megannicole51
  • megannicole51
yup im good!!! thank you! i just like working through these problems with people until i feel like ive got it:)

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