Chapter 10.7, Problem 29E

To determine

The partial decomposition of the given rational function is to be estimated:

  $\frac{2x}{4x^2+12x+9}$

Answer to Problem 29E

The partial decomposition of $\frac{2x}{4x^2+12x+9}$ is to be estimated as: $\frac{1}{\left(2x+3\right)}+\frac{-3}{{\left(2x+3\right)}^2}$

Explanation of Solution

Given:

  $\frac{2x}{4x^2+12x+9}$

Concept Used:

The process of starting the given rational number with simplified answer steps is the partial-fraction decomposition.

Calculation:

First, simply the given eq:

  $\frac{2x}{4x^2+12x+9}=\frac{2x}{{\left(2x+3\right)}^2}$

The form of the partial decomposition is expressed by the following expression:

  $\frac{2x}{{\left(2x+3\right)}^2}=\frac{A}{\left(2x+3\right)}+\frac{B}{{\left(2x+3\right)}^2}$

Now, write the right hand side of the equation into a single fraction:

  $\frac{2x}{{\left(2x+3\right)}^2}=\frac{B+\left(2x+3\right)A}{{\left(2x+3\right)}^2}$

As the denominators of both eqns. are equal so use the equality of numerators:

  $\begin{array}{l}2x=\left(2x+3\right)A+B\\ 2x=2xA+3A+B\end{array}$

The coefficients near the like terms should be equal, so the following system is obtained:

  $\begin{array}{l}2A=2\\ A=1\mathrm{...}\left(1\right)\\ 3A+B=0\mathrm{...}\left(2\right)\end{array}$

Now to get value of $B$ , put the value of $A=1$ in eq. (2):

  $\begin{array}{l}3A+B=0\\ 3+B=0\\ B=-3\end{array}$

Therefore, the partial decomposition of the given rational function is to be estimated as:

  $\begin{array}{l}\frac{2x}{{\left(2x+3\right)}^2}=\frac{A}{\left(2x+3\right)}+\frac{B}{{\left(2x+3\right)}^2}\\ \frac{2x}{{\left(2x+3\right)}^2}=\frac{1}{\left(2x+3\right)}+\frac{-3}{{\left(2x+3\right)}^2}\\ \frac{2x}{4x^2+12x+9}=\frac{1}{\left(2x+3\right)}+\frac{-3}{{\left(2x+3\right)}^2}\end{array}$

Conclusion:

Hence, $\frac{2x}{4x^2+12x+9}=\frac{1}{\left(2x+3\right)}+\frac{-3}{{\left(2x+3\right)}^2}$