Answer
\[\frac{7x^2-20}{(x-2)(x^2+4)}=\frac{1}{(x-2)}+\frac{6(x+2)}{x^2+4}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{7x^2-20}{(x-2)(x^2+4)}=\frac{A}{(x-2)}+\frac{Bx+C}{x^2+4}\;\;\;...(*)\]
\[7x^2-20=A(x^2+4)+(Bx+C)(x-2)\]
\[7x^2-20=A(x^2+4)+(Bx^2-2Bx+Cx-2C)\]
\[7x^2-20=(A+B)x^2+(C-2B)x+(4A-2C)\]
Comparing like coefficients
\[A+B=7\;\;\;...(1)\]
\[C-2B=0\Rightarrow C=2B\;\;\;...(2)\]
\[4A-2C=-20\;\;\;...(3)\]
Using (2) in (3) then (3) becomes
\[4A-4B=-20\Rightarrow A-B=-5\;\;\;...(4)\]
Add (1) and (4)
\[2A=2\Rightarrow A=1\]
From (1)
\[B=6\]
From (2)
\[C=12\]
From (*)
\[\frac{7x^2-20}{(x-2)(x^2+4)}=\frac{1}{(x-2)}+\frac{6x+12}{x^2+4}\]
\[\frac{7x^2-20}{(x-2)(x^2+4)}=\frac{1}{(x-2)}+\frac{6(x+2)}{x^2+4}\]
Hence,
\[\frac{7x^2-20}{(x-2)(x^2+4)}=\frac{1}{(x-2)}+\frac{6(x+2)}{x^2+4}.\]
