## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{2}{x+2}+\dfrac{1}{x-1}$
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: $$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\quad \quad(1)$$ Now, write the RHS with a common denominator using their $LCD$. \begin{align*} \frac{3x}{(x+2)(x-1)}&=\frac{A(x-1)+B(x+2)}{(x+2) (x-1)}\\ \\ \end{align*} Equate the numerators to obtain: $$3x= A(x-1)+B(x+2) \\ 3x=Ax-A+Bx+2B \\ 3x=Ax-A+Bx+2B \\ 3x=Ax+Bx-A+2B \\3x= (A+B)x+(-A+2B)$$ Equate the coefficients of the polynomials on the $LHS$ and $RHS$ to obtain: $$A+B=3\quad \quad(2) \\ -A+2B=0 \quad (3)$$ Add Equations $(2)$ and $(3)$ to obtain: $$(A+B)+(-A+2B)=3+0 \\3B=3 \\ B=1$$ Substitute $B=1$ into Equation $(3)$ to solve for $A$. $$-A+(2)(1)=0\\ -A+2=0 \\ A=2$$ Thus, equation $(1)$ becomes: $\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{2}{x+2}+\frac{1}{x-1}$