Answer:
[tex]\frac{x - 20}{ x^2 + x - 20} = \frac{5}{9(x-4)}+\frac{4}{(x+5)}[/tex]
Step-by-step explanation:
Given
[tex]\frac{x - 20}{ x^2 + x - 20}[/tex]
Required
Decompose into partial fraction
Start by factorizing the denominator:
[tex]x^2 + x - 20 =x^2 + 5x -4x- 20[/tex]
[tex]x^2 + x - 20 =x(x + 5)-4(x+5)[/tex]
[tex]x^2 + x - 20 =(x-4)(x+5)[/tex]
So, we have:
[tex]\frac{x - 20}{ x^2 + x - 20} = \frac{A}{x-4}+\frac{B}{x+5}[/tex]
Take LCM
[tex]\frac{x - 20}{ x^2 + x - 20} = \frac{A(x+5)+B(x-4)}{(x-4)(x+5)}[/tex]
Cancel out the denominator
[tex]x - 20 = A(x+5)+B(x-4)}[/tex]
Open brackets
[tex]x - 20 = Ax+5A+Bx-4B[/tex]
Collect Like Terms
[tex]x - 20 = Ax+Bx+5A-4B[/tex]
[tex]x - 20 = (A+B)x+5A-4B[/tex]
By comparison;
[tex](A +B)x = x[/tex] ==>[tex]A + B = 1[/tex]
[tex]5A - 4B = 1[/tex]
Make A the subject in [tex]A + B = 1[/tex]
[tex]A = 1 - B[/tex]
Substitute 1 - B for A in [tex]5A - 4B = 1[/tex]
[tex]5(1-B) -4B = 1[/tex]
[tex]5-5B -4B = 1[/tex]
[tex]5-9B = 1[/tex]
Collect Like Terms
[tex]-9B = 1 - 5[/tex]
[tex]-9B = - 4[/tex]
[tex]B = \frac{4}{9}[/tex]
[tex]A = 1 - B[/tex]
[tex]A = 1 - \frac{4}{9}[/tex]
[tex]A = \frac{9-4}{9}[/tex]
[tex]A = \frac{5}{9}[/tex]
So:
[tex]\frac{x - 20}{ x^2 + x - 20} = \frac{A}{x-4}+\frac{B}{x+5}[/tex]
[tex]\frac{x - 20}{ x^2 + x - 20} = \frac{5}{9(x-4)}+\frac{4}{(x+5)}[/tex]
