Answer
$\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{1}{x+2}+\frac{2}{x-4}$
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{A}{x+2}+\frac{B}{x-4}\quad \quad(1)$$
Now, write the RHS with a common denominator.
$$\displaystyle \frac{3x}{(x+2)(x-4)}=\dfrac{(x-4)A+(x+2)B}{(x+2)(x-4)}\\
\displaystyle \frac{3x}{(x+2)(x-4)}=\dfrac{Ax-4A+Bx+2B}{(x+2)(x-4)}\\
\displaystyle \frac{3x}{(x+2)(x-4)}=\dfrac{Ax+Bx+(-4A+2B)}{(x+2)(x-4)}\\
\displaystyle \frac{3x}{(x+2)(x-4)}=\dfrac{(A+B)x+(-4A+2B)}{(x+2)(x-4)}$$
Next, equate the numerators to obtain:
$$(A+B)x+(-4A+2B)=3x$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+B=3 ~~~(2) \\ -4A+2B=0 ~~~(3)$$
Equation (2) yields: $$A=3-B~~~~(4)$$
Plug $A=3-B$ into the Equation-(3) to solve for $B$.
Thus, we have: $$-4(3-B)+2B=0 \\-12+4B+2B=0\\-12+6B=0\\6B=12 \\ B=2$$
Finally, plug $B=2$ into equation (4) to obtain: $$A=3-2=1$$
Therefore, the equation (1) becomes: $\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{1}{x+2}+\frac{2}{x-4}$
