Answer
The simplified form of the expression $\frac{{{x}^{2}}+11}{3-x}$ for $x=4i$ is $-\frac{3}{5}-\frac{4}{5}i$.
Work Step by Step
Consider the expression, $\frac{{{x}^{2}}+11}{3-x}$
Substitute $x=4i$ in the expression $\frac{{{x}^{2}}+11}{3-x}$.
$\begin{align}
& \frac{{{x}^{2}}+11}{3-x}=\frac{{{\left( 4i \right)}^{2}}+11}{3-\left( 4i \right)} \\
& =\frac{16{{i}^{2}}+11}{3-4i}
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \frac{{{x}^{2}}+11}{3-x}=\frac{16\left( -1 \right)+11}{3-4i} \\
& =\frac{-16+11}{3-4i} \\
& =\frac{-5}{3-4i}
\end{align}$
Multiply by complex conjugate of the denominator in the numerator and the denominator.
\[\frac{{{x}^{2}}+11}{3-x}=\frac{-5}{\left( 3-4i \right)}\cdot \frac{\left( 3+4i \right)}{\left( 3+4i \right)}\]
Distribute $-5$ within the parentheses in numerator and use the FOIL method in the denominator.
\[\begin{align}
& \frac{{{x}^{2}}+11}{3-x}=\frac{-5\left( 3+4i \right)}{\left( 3-4i \right)\left( 3+4i \right)} \\
& =\frac{-15-20i}{9+12i-12i-16{{i}^{2}}} \\
& =\frac{-15-20i}{9-16{{i}^{2}}}
\end{align}\]
Replace the value ${{i}^{2}}=-1$.
\[\begin{align}
& \frac{{{x}^{2}}+11}{3-x}=\frac{-15-20i}{9-16\left( -1 \right)} \\
& =\frac{-15-20i}{25} \\
& =-\frac{15}{25}-\frac{20}{25}i \\
& =-\frac{3}{5}-\frac{4}{5}i
\end{align}\]
Therefore, the simplified form of the expression $\frac{{{x}^{2}}+11}{3-x}$ for $x=4i$ is $-\frac{3}{5}-\frac{4}{5}i$.
