Case 2. Repeated linear factor in denominator.

[latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2}[/latex]

[latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2} = \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}[/latex]

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

[latex s=1]\Big((x-1)(x-2)^2\Big) \Big(\frac{4x^2}{(x-1)(x-2)^2}\Big) = \Big((x-1)(x-2)^2\Big) \Big(\frac{A}{(x-1)}\Big)+ \Big((x-1)(x-2)^2\Big) \Big(\frac{B}{(x-2)}\Big)+ \Big((x-1)(x-2)^2\Big) \Big(\frac{C}{(x-2)^2}\Big)[/latex]

[latex s=1]4x^2 = A(x-2)^2+B(x-1)(x-2)+C(x-1)[/latex]  …………(1)

Now we have [latex] (x-1)[/latex] and [latex](x-2)^2[/latex]

[latex]x-1=0 \longmapsto x=1[/latex]

[latex](x-2)^2=0 \longmapsto x=2[/latex]

Put [latex]\textbf{x=1}[/latex] and [latex]\textbf{x=2}[/latex] in equation (1) one by one

At [latex]\textbf{x=1}[/latex]



At [latex]\textbf{x=2}[/latex]



However, there is no value of x that will allow us to eliminate the first and third term leaving only the middle term that we can use to solve B. While this may appear to be a problem, it actually isn’t it. At this point we know two of the three constants. So all we need to do is chose any other value of x that would be easy to work with, x = 0 seems useful here.

At [latex]\textbf{x=0}[/latex]



[latex s=2]F(x) = \frac{8x-42}{x^2+3x-18} = \frac{8x-42}{(x+6)(x-3)} = \frac{4}{x+6}+\frac{0}{x-3}+\frac{16}{x-3}[/latex]

[latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2} = \frac{4}{(x-1)}+\frac{0}{(x-2)}+\frac{16}{(x-2)^2}[/latex]

[latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2} = \frac{4}{(x-1)}+\frac{16}{(x-2)^2}[/latex]


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