Question

Write the partial fraction decomposition of the rational expression.

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

Short answer

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

$\frac{4}{5\left(2x-1\right)}+\frac{8}{5\left(x+2\right)}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

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

Step 2. Decompose the rational expression.

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

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

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

Where, Aand B are to be determined.

Multiply every side by $\left(x+2\right)\left(2x-1\right)$,

$4x=A\left(2x-1\right)+B\left(x+2\right)......\left(ii\right)$

Step 3. Find the value of A and B.

Assume$x=-2$in equation (ii),

$$\begin{gathered} 4\left(-2\right)=A\left(2\left(-2\right)-1\right)+B\left(\left(-2\right)+2\right) \\ -8=A\left(-5\right) \\ A=\frac{8}{5} \end{gathered}$$

Assume $x=0$in equation (ii),

role="math" localid="1647593570391" $$\begin{gathered} 4\left(0\right)=A\left(2\left(0\right)-1\right)+B\left(\left(0\right)+2\right) \\ 0=-A+2B \\ B=\frac{A}{2}......\left(iii\right) \end{gathered}$$

Substitute the value of Ain equation (iii),

$$\begin{gathered} B=\frac{{\displaystyle \frac{8}{5}}}{2} \\ B=\frac{4}{5} \end{gathered}$$

Step 4. Substitute the values in equation (i).

On substituting the values of A and Bin equation (i),

$$\begin{gathered} \frac{4x}{2x^2+3x-2}=\frac{{\displaystyle \frac{4}{5}}}{2x-1}+\frac{{\displaystyle \frac{8}{5}}}{x+2} \\ \frac{4x}{2x^2+3x-2}=\frac{4}{5\left(2x-1\right)}+\frac{8}{5\left(x+2\right)} \end{gathered}$$