Case 3. Quadratic term in denominator.

[latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)}[/latex]

[latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)} = \frac{A}{(x)}+\frac{Bx+C}{(x^2+2x-6)}[/latex]

Multiply by  [latex s=1]x(x^2+2x-6) [/latex] with above equation on both sides we get

[latex s=1]\Big(x(x^2+2x-6)\Big) \Big(\frac{8x^2-12}{x(x^2+2x-6)}\Big) = \Big(x(x^2+2x-6)\Big) \Big(\frac{A}{(x)}\Big)+ \Big(x(x^2+2x-6)\Big) \Big(\frac{B}{x(x^2+2x-6)}\Big)[/latex]

[latex]8x^2-12 = A(x^2+2x-6)+x(Bx+C)[/latex]

We could choose [latex]x=0[/latex] to get the value of A, but that’s the only constant that we could get using this method and so it just won’t work all that well here.So, multiply the right side out and then collect all the like terms as follow

[latex]8x^2-12 = Ax^2+2Ax-6A+Bx^2+Cx[/latex]

[latex]8x^2-12 = (A+B)x^2+(2A+C)x-6A[/latex]

Now comparing the coefficient

[latex]-6A=-12 \longmapsto \framebox[1.1\width]{A=2}[/latex]

[latex]2A+C=0 \longmapsto \framebox[1.1\width]{C=-4}[/latex]

[latex]A+B=8 \longmapsto \framebox[1.1\width]{B=6}[/latex]

[latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)} = \frac{A}{(x)}+\frac{Bx+C}{(x^2+2x-6)}[/latex]

[latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)} = \frac{2}{(x)}+\frac{6x+4}{(x^2+2x-6)}[/latex]

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