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

$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{-\frac{1}{12}x-\frac{1}{3}}{x^2+2x+4}$
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: $$\displaystyle \frac{1}{(x-2)(x^2+2x+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4} \quad \quad(1)$$ Now, write the RHS with a common denominator. \begin{align*}\displaystyle \frac{1}{(x-2)(x^2+2x+4)}&=\frac{A(x^2+2x+4)+(x-2)(Bx+C)}{(x-2)(x^2+2x+4)}\\ \\\frac{1}{(x-2)(x^2+2x+4)}&=\frac{Ax^2+2Ax+4A+Bx^2-2Bx+Cx-2C}{(x-2)(x^2+2x+4)} \\\frac{1}{(x-2)(x^2+2x+4)}&=\frac{(A+B)x^2+(2A-2B+C)x+(4A-2C)}{(x-2)(x^2+2x+4)} \end{align*} Next, equate the numerators to obtain: $$(A+B)x^2+(2A-2B+C)x+(4A-2C)=1$$ Equate the coefficients of the polynomials on the LHS and RHS to obtain: $$A+B=0 \quad \quad\ (2) \\ 2A-2B+C=0 \quad \quad (3) \\4A-2C=1\quad \quad(4)$$ Equation $(2)$ yields: $A=-B$. Plug $A=-B$ into the Equation $(3)$. $$2(-B)-2B+C=1\\ -2B-2B+C=0\\ -4B+C=0 \\ C=4B ~~~~(5)$$ Next, plug $A=-B$ and $C=4B$ into the Equation $(4)$. $$4(-B)-2(4B)=1\\ -4B-8B=1\\ -12B=1 \\ B=\dfrac{-1}{12}$$ Hence, $A=-B=-(\dfrac{-1}{12})=\dfrac{1}{12}$ Finally, plug $B=-\frac{1}{12}$ into Equation $(5)$ to solve for $C$. $$C =4(\dfrac{-1}{12})=\dfrac{-1}{3}$$ Therefore, equation $(1)$ becomes: $$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{\frac{-1}{12}x+\dfrac{-1}{3}}{x^2+2x+4}$$ or, $$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{-\frac{1}{12}x-\frac{1}{3}}{x^2+2x+4}$$