Q.22 - Precalculus Enhanced with Graphing Utilities

Question

Write the partial fraction decomposition of each rational expression.

$\frac{2x+4}{x^3-1}$

Step-by-Step Solution

Answer

The partial fraction decomposition is,$\frac{2}{x-1}+\frac{-2(x+1)}{x^2+x+1}$.

Step 1. Given Information.

The rational expression is,

$\frac{2x+4}{x^3-1}$.

Step 2. Finding the values.

Factoring the rational expression,

$\frac{2x+4}{x^3-1}=\frac{2x+4}{(x-1)(x^2+x+1)}$

Decomposition of rational expressions,

$\frac{2x+4}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}........\left(1\right)$

Multiplying both sides by $(x-1)(x^2+x+1)$,

$$\begin{gathered} 2x+4=A(x^2+x+1)+(Bx+C)(x-1) \\ 2x+4=A{x}^2+Ax+A+B{x}^2-Bx+Cx-C \\ 2x+4=(A+B)x^2+(A-B+C)x+(A-C)............\left(2\right) \end{gathered}$$

Equating the coefficients of the like powers of $x$, we get,

$$\begin{gathered} A+B=0........\left(3\right) \\ A-B+C=2........\left(4\right) \\ A-C=4.........\left(5\right) \end{gathered}$$

From equation (3), we get, $B=-A$

Inputting the value of $B$, in equation (4), we get,

$2A+C=2.......\left(6\right)$

Solving the equation (5) and (6)we get,

$$\begin{gathered} A=2 \\ C=-2 \\ B=-2 \end{gathered}$$

Step 3. Partial fraction decomposition

The partial fraction decomposition is,

$\frac{2x+4}{x^3-1}=\frac{2}{x-1}+\frac{-2x-2}{x^2+x+1}$

             $=\frac{2}{x-1}+\frac{-2(x+1)}{x^2+x+1}$

Step 4. Checking the solution.

Adding the rational expressions,

$\frac{2}{x-1}+\frac{-2(x+1)}{x^2+x+1}=\frac{2(x^2+x+1)+(-2(x+1\left)\right(x-1)}{(x-1)(x^2+x+1)}$

                                $$\begin{gathered} =\frac{2x^2+2x+2-2x^2+2}{x^3-1} \\ =\frac{2x+4}{x^3-1} \end{gathered}$$

The solution is true.