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As the title says everything.

 Jul 29, 2017

Best Answer 

 #1
avatar+95017 
+3

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 :)

 Jul 29, 2017
 #1
avatar+95017 
+3
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

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