MHB Find the partial fraction decomposition for the rational function.

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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 the numerators, the equation \(-4x^2 - 8x - 19=(A+C)x^2+(B-9A)x+(2C-9B)\) is derived, leading to a system of equations for \(A\), \(B\), and \(C\). Solving this system provides the values needed to complete the partial fraction decomposition. Ultimately, substituting these values back into the original decomposition format yields the final result.
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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