Q. 47 - Precalculus Enhanced with Graphing Utilities

Question

Use the division algorithm to rewrite the improper fraction as the sum of a quotient and proper fraction. Find the partial fraction decomposition of the proper fraction. Finally, express the improper fraction as the sum of a quotient and the partial fraction decomposition.

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

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression is given below,

$\frac{{\displaystyle \frac{51}{5}}}{x+4}+\frac{-{\displaystyle \frac{1}{5}}}{x-1}$

On expression the improper fraction as the sum of a quotient and the partial fraction decomposition,

$x-2+\frac{10x-11}{x^2+3x-4}$

Step 1. Given information.

Consider the given question,

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

Step 2. Use the division algorithm.

Using the division algorithm,

Therefore, it can be said,

$\frac{x^3+x^2-3}{x^2+3x-4}=x-2+\frac{10x-11}{x^2+3x-4}$.

Step 3. Factor the denominator.

The denominator can be factorized as $x^2+3x-4=\left(x+4\right)\left(x-1\right)$.

The denominator has a repeated non-reducible quadratic factors. Then we decompose the rational expression as follows,

$$\begin{gathered} \frac{10x-11}{x^2+3x-4}=\frac{10x-11}{\left(x+4\right)\left(x-1\right)} \\ \frac{10x-11}{x^2+3x-4}=\frac{A}{x+4}+\frac{B}{x-1} ...... \left(i\right) \end{gathered}$$

Then,

$$\begin{gathered} 10x-11=A\left(x-1\right)+B\left(x+4\right) \\ 10x-11=Ax-A+Bx+4B \\ 10x-11=\left(A+B\right)x+\left(-A+4B\right) \end{gathered}$$

Therefore, it can be written,

$$\begin{gathered} A+B=10 ...... \left(ii\right) \\ -A+4B=-11 ...... \left(iii\right) \end{gathered}$$

Step 4. Add equations (ii) and (iii).

On adding equations (ii) and (iii),

$$\begin{gathered} \left(A+B\right)+\left(-A+4B\right)=10-11 \\ 5B=-1 \\ B=-\frac{1}{5} \end{gathered}$$

Substitute the value of B in equation (ii),

$$\begin{gathered} A-\frac{1}{5}=10 \\ A=\frac{51}{5} \end{gathered}$$

Substitute the values in equation (i),

$\frac{10x-11}{x^2+3x-4}=\frac{{\displaystyle \frac{51}{5}}}{x+4}+\frac{-{\displaystyle \frac{1}{5}}}{x-1}$