Evaluate the expression $\displaystyle \frac{(1+2i)(3-i)}{2+i}$ in the form of $a + bi$.
$
\begin{equation}
\begin{aligned}
&= \frac{(1+2i)(3-i)}{2+i}\\
\\
&= \frac{3-i+6i-2i^2}{2+i} && \text{Use FOIL method}\\
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&= \frac{3+5i -2 (-1)}{2+i} && \text{recall that } i^2 = -1\\
\\
&= \frac{5+5i}{2+i} && \text{Simplify the numerator}\\
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&= \left( \frac{5+5i}{2+i} \right) \left( \frac{2-i}{2-i} \right) && \text{Multiply by the denominator of the conjugate}\\
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&= \frac{10-5i+10i-5i^2}{2^2 - i^2} && \text{Apply FOIL method}\\
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&= \frac{10-5i+10i-5(-1)}{4-(-1)} && \text{recall that } i^2 = -1\\
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&= \frac{15 + 5i}{5} && \text{Simplify and group terms}\\
\\
&= 3 + i
\end{aligned}
\end{equation}
$
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