Splitting into partial fractions

[Mathme]

Hi, I'm having a hard time understanding how my book does the split of 1/(x-2)(x-5)

Can anyone help me with the logic behind ending up with (-1/3)/(x-2) + (1/3)/(x-5)?

Thanks!

 

[tkhunny]

This should be in your text book. It requires some algebra manipulation skills.

The expression c/(A*B) has an equivalent expression a/A + b/B for some values of 'a' and 'b' that are not necessarily obvious. It is easy enough to demonstrate by working backwards.

Most simply stated, it is just Algebra 1 Addition of Fractions in Reverse.

 

[soroban]

Hello, Mathme!

You're wondering how this is done.
You should been shown the method before they threw this problem at you.

\(\displaystyle \L\frac{1}{(x\,-\,2)(x\,-\,5)}\;=\;\frac{-\frac{1}{3}}{x\,-\,2}\,+\,\frac{\frac{1}{3}}{x\,-\,5}\)

We conjecture that the fraction can be separated into two "partial fractions":

\(\displaystyle \L\;\;\;\frac{1}{(x\,-\,2)(x\,-\,5)}\;=\;\frac{A}{x\,-\,2} \,+ \,\frac{B}{x\,-\,5)\)


Multiply through by the LCD: \(\displaystyle \,1\;=\;(x\,-\,5)A \,+\,(x\,-\,2)B\)


Now we select some "good' values for \(\displaystyle x\).

Let \(\displaystyle x\,=\,2:\;\;1\;=\;(-3)A + (0)B\;\;\Rightarrow\;\;-3A\,=\,1\;\;\Rightarrow\;\;A\,=\,-\frac{1}{3}\)

Let \(\displaystyle x\,=\,5:\;\;1\;=\;(0)A\,+\,(3)B\;\;\Rightarrow\;\;3B\,=\,1\;\;\Rightarrow\;\;B\,=\,\frac{1}{3}\)

And we've got it! . . .
\(\displaystyle \L\;\;\;\frac{1}{(x\,-\,2)(x\,-\,5)}\:=\;\frac{-\frac{1}{3}}{x\,-\,2} \,+\,\frac{\frac{1}{3}}{x\,-\,5} \;=\;\frac{1}{3}\left(\frac{-1}{x\,-\,2}\,+\,\frac{1}{x\,-\,5}\right)\)

 

[stapel]

Please take the time to study the method provided by Soroban. It isn't the method I've commonly seen in textbooks (so your book may use something else, such as systems of equations), but -- at least in my opinion -- his method makes solving for A, B, etc, a lot faster and easier.

Eliz.