Partial Fractions

**louboutinlover** · April 30th 2009, 12:10 PM

$s = \frac{s^2+8}{(s^2+4)(s^2+9)}$

$A(s^2+9) + B(s^2+4) = s^2+8$

Uncertain of what number to substitute to solve ?

**e^(i*pi)** · April 30th 2009, 12:15 PM

Originally Posted by louboutinlover

$s = \frac{s^2+8}{(s^2+4)(s^2+9)}$

$A(s^2+9) + B(s^2+4) = s^2+8$

Uncertain of what number to substitute to solve ?

You can make some equations by comparing coefficients

**louboutinlover** · April 30th 2009, 01:09 PM

how do you compare coefficients ?

**HallsofIvy** · April 30th 2009, 01:56 PM

Originally Posted by louboutinlover

$s = \frac{s^2+8}{(s^2+4)(s^2+9)}$

$A(s^2+9) + B(s^2+4) = s^2+8$

Uncertain of what number to substitute to solve ?

ANY numbers will do! Putting any 2 numbers in for s will give two equations for A and B. For example, if s= 0, A(0+ 9)+ B(0+ 4)= 0+ 8 or 9A+ B= 8. If s= 1, A(1+ 9)+ B(1+ 4)= 1+ 8 or 10A+ 5B= 9.

**HallsofIvy** · April 30th 2009, 02:01 PM

Originally Posted by louboutinlover

how do you compare coefficients ?

You had $A(s^2+9) + B(s^2+4) = s^2+8$ . Multiply it out: $As^2+ 9A+ Bs^2+ 4B= s^2+ 8$
$(A+ B)s^2+ (9A+ 4B)= s^2+ 8$
The coefficient of $s^2$ is A+ B on one side, 1 on the other: you must have A+ B= 1. The coefficient of s is 0 on both sides (whew!). The constant term is 9A+ 4B on one side and 8 on the other. You must have 9A+ 4B= 8.

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