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 [latex]\textbf(x+6)(x-3)[/latex] 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]

 

[latex]8x-42=A(x-3)+B(x+6)[/latex]  …………(1)

Now we have [latex]\textbf{(x+6)}[/latex] and [latex]\textbf{(x-3)}[/latex]

[latex]x+6=0 \longmapsto x=-6[/latex]

[latex]x-3=0 \longmapsto x=3[/latex]

Put [latex]\textbf{x=-6}[/latex] and [latex]\textbf{x=3}[/latex] in equation (1) one by one

At [latex]\textbf{x=-6}[/latex]

[latex]8(-6)-42=A(-6-3)+B(-6+6)[/latex]

[latex]-48-12=-9A[/latex]

[latex]-90=-9A[/latex]

[latex]\framebox[1.1\width]{A = 10}[/latex]

At [latex]\textbf{x=3}[/latex]

[latex]8(3)-42=A(3-3)+B(3+6)[/latex]

[latex]24-42=9B[/latex]

[latex]-18=9B[/latex]

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

Now we got [latex]\textbf A[/latex] and [latex]\textbf B[/latex] put back

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