The problem asks to decompose the rational function $\frac{8x-12}{x^2+3x-18}$ into partial fractions.

1. Problem Description

The problem asks to decompose the rational function $\frac{8x-12}{x^2+3x-18}$ into partial fractions.

2. Solution Steps

First, we factor the denominator:

$x^2+3x-18 = (x+6)(x-3)$

Now we can write the partial fraction decomposition as:

$\frac{8x-12}{(x+6)(x-3)} = \frac{A}{x+6} + \frac{B}{x-3}$

To find A and B, we multiply both sides by $(x+6)(x-3)$:

$8x-12 = A(x-3) + B(x+6)$

We can solve for A and B by choosing convenient values for x.

Let $x = 3$:

$8(3) - 12 = A(3-3) + B(3+6)$

$24 - 12 = 0 + 9B$

$12 = 9B$

$B = \frac{12}{9} = \frac{4}{3}$

Let $x = -6$:

$8(-6) - 12 = A(-6-3) + B(-6+6)$

$-48 - 12 = -9A + 0$

$-60 = -9A$

$A = \frac{-60}{-9} = \frac{20}{3}$

Thus, the partial fraction decomposition is:

$\frac{8x-12}{x^2+3x-18} = \frac{\frac{20}{3}}{x+6} + \frac{\frac{4}{3}}{x-3} = \frac{20}{3(x+6)} + \frac{4}{3(x-3)}$

3. Final Answer

$\frac{20}{3(x+6)} + \frac{4}{3(x-3)}$