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

$\dfrac{-4}{x}+\dfrac{4}{x-1}$
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated linear factors. So, we will write the given rational expression into partial fraction form as: $\displaystyle \frac{4}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$ Now, write the RHS with a common denominator. $\displaystyle \frac{4}{x(x-1)}=\frac{A(x-1)+Bx}{x(x-1}\quad \quad(1)$ Next, equate the numerators to obtain: $Ax-A+Bx=4 \\(Ax+Bx)-A=4 \\(A+B)x-A=4$ Equate the coefficients of the polynomials on the LHS and RHS to obtain: $$-A =4 \implies A=-4$$ Also. $$\\A+B=0 \quad \quad(2)$$ Plug $A=-4$ into the Equation $(2)$ to solve for $B$. $$-4+B=0\implies B=4$$ Thus, the equation $(1)$ becomes: $$\displaystyle \frac{4}{x(x-1)}=\frac{-4}{x}+\frac{4}{x-1}$$