10.38 Form the polynomial of the least degree with real coefficients that has roots $x_{1}=-2$ and $x_{2}=4-3 i$.

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10.38 Since the desired equation must have real coefficients, the number of its complex roots is even. Considering the existence of another complex root $x_{3}=4+3 i$, we conclude that the equation of the lowest degree is a cubic equation:

$$
\begin{aligned}
& (x+2)(x-4+3 i)(x-4-3 i)=0 ;(x+2)\left((x-4)^{2}+9\right)=0 \\
& (x+2)\left(x^{2}-8 x+25\right)=0 ; x^{3}-6 x^{2}+9 x+50=0
\end{aligned}
$$

Answer: $x^{3}-6 x^{2}+9 x+50=0$.