Case 1. Distinct linear factor in denominator

Let us see the how to slove when distinct linear factor is in denominator.

Distinct Linear Factor:

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

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

Multiply by $\textbf(x+6)(x-3)$ on both sides with above equation we get

[latex s=-1]\Big((x+6)(x-3)\Big)\Big(\frac{8x-42}{x^2+3x-18}\Big) = \Big(\Big((x+6)(x-3)\Big)\Big(\frac{A}{x+6}\Big)\Big) + \Big(\Big((x+6)(x-3)\Big)\Big(\frac{B}{x-3}\Big)\Big)[/latex]

$8x-42=A(x-3)+B(x+6)$  …………(1)

Now we have $\textbf{(x+6)}$ and $\textbf{(x-3)}$

$x+6=0 \longmapsto x=-6$

$x-3=0 \longmapsto x=3$

Put $\textbf{x=-6}$ and $\textbf{x=3}$ in equation (1) one by one

At $\textbf{x=-6}$

$8(-6)-42=A(-6-3)+B(-6+6)$

$-48-12=-9A$

$-90=-9A$

$\framebox[1.1\width]{A = 10}$

At $\textbf{x=3}$

$8(3)-42=A(3-3)+B(3+6)$

$24-42=9B$

$-18=9B$

$\framebox[1.1\width]{B=-2}$

Now we got $\textbf A$ and $\textbf B$ put back

[latex s=2]F(x) = \frac{8x-42}{x^2+3x-18} = \frac{10}{x+6}+\frac{-2}{x-3}[/latex]

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