+0  
 
0
68
1
avatar+6765 

As the title says everything.

MaxWong  Jul 29, 2017

Best Answer 

 #1
avatar+89839 
+2

How do I do partial fractions.

 

Hi Max :)

 

Here is an example.

 

\(\frac{4}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A}{(x+2)(x-1)}+\frac{Bx+2B}{(x+2)(x-1)}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A+Bx+2B}{(x+2)(x-1)}\\ Ax-A+Bx+2B=4\\ (A+B)x+2B-A=4\\ A+B=0 \qquad and \qquad 2B-A=4\\ A=-B \qquad and \qquad 2B-A=4\\ 2B--B=4\\ 3B=4\\ B=\frac{4}{3}\qquad and \qquad A=-\frac{4}{3}\\ so\\ \frac{4}{(x+2)(x-1)}=\frac{-4/3}{x+2}+\frac{4/3}{x-1}\\ \)

 

There are a number of you tube clips on this topic that will probably help more :)

Melody  Jul 29, 2017
Sort: 

1+0 Answers

 #1
avatar+89839 
+2
Best Answer

How do I do partial fractions.

 

Hi Max :)

 

Here is an example.

 

\(\frac{4}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A}{(x+2)(x-1)}+\frac{Bx+2B}{(x+2)(x-1)}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A+Bx+2B}{(x+2)(x-1)}\\ Ax-A+Bx+2B=4\\ (A+B)x+2B-A=4\\ A+B=0 \qquad and \qquad 2B-A=4\\ A=-B \qquad and \qquad 2B-A=4\\ 2B--B=4\\ 3B=4\\ B=\frac{4}{3}\qquad and \qquad A=-\frac{4}{3}\\ so\\ \frac{4}{(x+2)(x-1)}=\frac{-4/3}{x+2}+\frac{4/3}{x-1}\\ \)

 

There are a number of you tube clips on this topic that will probably help more :)

Melody  Jul 29, 2017

14 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details