Find the partial fraction decomposition for the rational function.

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SUMMARY

The discussion focuses on finding the partial fraction decomposition of the rational function $$\frac{-4x^2 - 8x - 19}{(x^2 + 2)(x-9)}$$. The decomposition is expressed as $$\frac{Ax+B}{x^2+2}+\frac{C}{x-9}$$. By equating coefficients, the system of equations derived is: $$-4=A+C$$, $$-8=B-9A$$, and $$-19=2C-9B$$. Solving this system yields the values for A, B, and C, which can then be substituted back into the original decomposition formula.

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shamieh
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Find the partial fraction decomposition for the rational function.

$$\frac{-4x^2 - 8x - 19}{(x^2 + 2)(x-9)}$$

I'm not sure what to do.
 
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shamieh said:
Find the partial fraction decomposition for the rational function.

$$\frac{-4x^2 - 8x - 19}{(x^2 + 2)(x-9)}$$

I'm not sure what to do.

$$\frac{-4x^2 - 8x - 19}{(x^2 + 2)(x-9)}=\frac{Ax+B}{x^2+2}+\frac{C}{x-9}$$

$$\frac{-4x^2 - 8x - 19}{(x^2 + 2)(x-9)}=\frac{(Ax+B)(x-9)+C(x^2+2)}{(x^2+2)(x-9)}$$

$$-4x^2 - 8x - 19=(Ax+B)(x-9)+C(x^2+2) \Rightarrow \\ -4x^2 - 8x - 19=Ax^2-9Ax+Bx-9B+Cx^2+2C \Rightarrow \\ -4x^2 - 8x - 19=(A+C)x^2+(B-9A)x+(2C-9B)$$

So $$-4=A+C, \ \ \ -8=B-9A, \ \ \ -19=2C-9B$$

Solving this system you will find the values of $A,B,C$.

Then substistute these values at $$\frac{Ax+B}{x^2+2}+\frac{C}{x-9}$$
 

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