Question

split into partial fractions....((5)/(y^3-25y))

Answer 1

https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/partial-fraction-expansion/v/partial-fraction-expansion-1

Answer 2

\[\Large \frac{5}{y^3-25y} \quad=\quad \frac{5}{y(y^2-5^2)}\]This might be a good first step right? :)
Looks like we can further factor though.
Remember how to break down the `difference of squares` ?

Answer 3

(A/y)+(B/y-5)+(C/(y-5)^2)?

Answer 4

Woops you may have factored your denominator incorrectly, let's see.

Answer 5

\[\Large \frac{5}{y(y^2-5^2)} \ne\frac{5}{y(y-5)^2}\]

Answer 6

thats how my professor broke down a problem in my notes....probably applies to a different type of problem

Answer 7

yah probably D:

Answer 8

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\[\Large y^2-25^2\]

Answer 9

woops* \(\Large y^2-5^2\)

Answer 10

im confused....why are you doing that?

Answer 11

watch the vid if you dont get it by the first vid there are 2 more vids :)

Answer 12

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.

Answer 13

@pgpilot326

Answer 14

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:
\[\frac{5}{y(y + 5)(y - 5)} = \frac{A}{y - 5} + \frac{B}{y} + \frac{C}{y + 5}\]

Answer 15

i put that earlier and someone said it was wrong?

Answer 16

actually my numerators were a bit different

Answer 17

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!

Answer 18

Next multiply both sides by \(y(y + 5)(y - 5))\), then simplify:

\[5 = y(y + 5)A + (y - 5)(y + 5)B + y(y - 5)C\]

Answer 19

why did you set the denominators up the way u did?

Answer 20

@megannicole51, because that's the form you want when expanding to partial fractions.

Answer 21

Now you have to expand the right side

Answer 22

The right side expands to:
\(y(y + 5)A + (y + 5)(y - 5)B + y(y - 5)C\)

\(= (y^2 + 5y)A + (y^2 - 25)B + (y^2 - 5y)C\)

\(=Ay^2 + A5y + By^2 - 25B + Cy^2 - C5y\)

\(=Ay^2 + By^2 + Cy^2 + A5y - C5y - 25B\)

\(=(A + B + C)y^2 + (A - C)5y - 25B\)

Answer 23

\( = (A + B + C)y^2 + (5A - 5C)y - 25B\)

Answer 24

i have to go to class but ill be back to take a look at ur expansion!

Answer 25

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:

\(0 = A + B + C\)
\(0 = 5A - 5C\)
\(5 = -25B\)

Answer 26

Then after solving the system you end up with:

\[A = \frac{1}{10}\]\[B = -\frac{1}{5}\]\[C = \frac{1}{10}\]

Answer 27

And the final form you get should be:

\[\frac{5}{y(y + 5)(y - 5)} = \frac{1}{10(y - 5)} -\frac{1}{5y} + \frac{1}{10(y + 5)}\]

Answer 28

0=A+B+C
0=5A−5C
5=−25B

how do you know this?

Answer 29

@Hero

Answer 30

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

Answer 31

oh that is not what you asked.

Answer 32

my question is just how do u know what those equal?

0=A+B+C
0=5A−5C
5=−25B

Answer 33

so you had:
5 = (A+B+C)y^2 + (5A-5C)y -25B

Answer 34

as you can see at the left side you have only a number (independent of y)

Answer 35

so both of the terms that multiply y and y^2 at the right side must be 0

Answer 36

A+B+C = 0
and
5A-5C = 0

Answer 37

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

Answer 38

oooh okay

Answer 39

Thank you to everyone who helped!

Answer 40

Sorry, I stepped away from my computer for a moment. @megannicole51, do you get everything else?

Answer 41

yup im good!!! thank you! i just like working through these problems with people until i feel like ive got it:)