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April 25th, 2014, 12:21 PM  #1 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1  Show that (fractions)
Show that (x+5)/((x1)(x+2)) can be written as 2/(x1)1/(x+2). Is there a specific tactic to approaching these questions or is it really straightforward? Last edited by skipjack; April 25th, 2014 at 01:52 PM. 
April 25th, 2014, 01:37 PM  #2 
Senior Member Joined: Feb 2010 Posts: 708 Thanks: 142 
I believe you have a typo on the left hand side. The problem should read $\displaystyle \dfrac{x+5}{(x1)(x+2)} = \dfrac{2}{x1}\dfrac{1}{x+2}$. What you want to do is called partial fractions. You can look this up online or in any reasonably good calculus book. 
April 25th, 2014, 01:47 PM  #3 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1 
Thank you sir, and no my book appears not to have this. Then again, it's not focused on calculus
Last edited by The; April 25th, 2014 at 01:48 PM. Reason: forgot how to english 
April 25th, 2014, 02:30 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162 
I've corrected the errors in the original equation. For information about partial fraction decomposition, follow the link or do a web search for easier articles about partial fractions. The simplest decompositions are easily obtained in one step (with the aid of some mental arithmetic) by use of the coverup rule. Last edited by skipjack; April 25th, 2014 at 04:29 PM. 
April 25th, 2014, 03:35 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
Essentially, for partial fraction decomposition you write down the result with a number of unknown values and then solve a system of simultaneous equations for the solution. In this case, we will write $$\frac{x+5}{(x1)(x+2)} = \frac{A}{x1}+\frac{B}{x+2}$$ And then we multiply by the denominator on the left. $$x+5=A(x+2) + B(x1)$$ Now we compare coefficients of the different powers of $x$. \begin{align*} 5 &= 2A  B &\text{from the units}\\ 1 &= A + B &\text{from the coefficients of $x$} \end{align*} Thus we have two equations for two unknowns which we can therefore solve. I encourage you to read more on the subject as it gets (a tiny bit) more complicated, but that approach will work for any denominator with distinct roots all of which are real (as long as the numerator is of lower order than the denominator). 

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