Answer
The standard form of the expression $\frac{4}{\left( 2+i \right)\left( 3-i \right)}$ is \[\frac{14}{25}-\frac{2}{25}i\].
Work Step by Step
Consider the expression, $\frac{4}{\left( 2+i \right)\left( 3-i \right)}$
Use the FOIL method.
$\begin{align}
& \frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{4}{6-2i+3i-{{i}^{2}}} \\
& =\frac{4}{6+i-{{i}^{2}}}
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{4}{6+i-\left( -1 \right)} \\
& =\frac{4}{7+i}
\end{align}$
Multiply by the complex conjugate of the denominator in the numerator and the denominator.
$\frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{4}{\left( 7+i \right)}\cdot \frac{\left( 7-i \right)}{\left( 7-i \right)}$
Distribute $4$ within the parentheses in the numerator and use the FOIL method in the denominator.
$\begin{align}
& \frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{4\left( 7-i \right)}{\left( 7+i \right)\left( 7-i \right)} \\
& =\frac{28-4i}{49-7i+7i-{{i}^{2}}} \\
& =\frac{28-4i}{49-{{i}^{2}}}
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{28-4i}{49-\left( -1 \right)} \\
& =\frac{28-4i}{49+1}
\end{align}$
Express the complex number in the standard form.
$\begin{align}
& \frac{4}{\left( 2+i \right)\left( 3-i \right)}=\frac{28-4i}{50} \\
& =\frac{28}{50}-\frac{4}{50}i \\
& =\frac{14}{25}-\frac{2}{25}i
\end{align}$
Therefore, the standard form of the expression $\frac{4}{\left( 2+i \right)\left( 3-i \right)}$ is $\frac{14}{25}-\frac{2}{25}i$.
