Case 4. Repeated quadratic term in denominator

[latex s=2]F(x) = \frac{2x^3+5x}{(x^2+3)^2}[/latex] [latex s=3]F(x) = \frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+3)^2}[/latex] Multiply an divide [latex](x^2+3)^2[/latex] with [latex]Ax+B [/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)(x^2+3)}+ \frac{Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)^2}+ \frac{Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)+Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3Ax+3B+Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3(A+C)x+(3B+D)}{(x^2+3)^2}[/latex] Comparing the coefficient we get [latex]\framebox[1.1\width]{A=2}[/latex] [latex]\framebox[1.1\width]{B=0}[/latex] [latex]3A+C=5 \longmapsto \framebox[1.1\width]{C=-1}[/latex] [latex]3B+D=0 \longmapsto \framebox[1.1\width]{D=0}[/latex] Put the value […]

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 […]

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) […]

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 […]