## Precalculus (6th Edition) Blitzer

The standard form of the expression $\frac{4}{\left( 2+i \right)\left( 3-i \right)}$ is $\frac{14}{25}-\frac{2}{25}i$.
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 $\frac{14}{25}-\frac{2}{25}i$.