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# How do I do partial fractions??

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As the title says everything.

MaxWong  Jul 29, 2017

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+91451
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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
+91451
+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 :)

Melody  Jul 29, 2017

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