# 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 $(x-1)(x-2)^2$ 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 $(x-1)$ and $(x-2)^2$

$x-1=0 \longmapsto x=1$

$(x-2)^2=0 \longmapsto x=2$

Put $\textbf{x=1}$ and $\textbf{x=2}$ in equation (1) one by one

At $\textbf{x=1}$

$4(1)^2=A(1-2)^2+B(1-1)(1-2)+C(1-1)$

$\framebox[1.1\width]{A=4}$

At $\textbf{x=2}$

$4(2)^2=A(2-2)^2+B(2-1)(2-2)+C(2-1)$

$\framebox[1.1\width]{C=16}$

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 $\textbf{x=0}$

$4(0)^2=A(0-2)^2+B(0-1)(0-2)+C(0-1)$

$\framebox[1.1\width]{B=0}$

[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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