jmc

Solution #1

Since $p(x)$ has real coefficients and has $3-2i$ as a root, it also has the complex conjugate, $3+2i$, as a root. The quadratic that has $3-2i$ and $3+2i$ as roots is $$\begin{array}{rl}\left(x-(3-2i)\right)\left(x-(3+2i)\right)& =(x-3+2i)(x-3-2i)\\ & =(x-3{)}^2-(2i{)}^2\\ & =x^2-6x+9+4\\ & =x^2-6x+13.\end{array}$$By the Factor Theorem, we know that $x^2-6x+13$ divides $p(x)$. Since $p(x)$ is cubic, it has one more root $r$. We can now write $p(x)$ in the form $$p(x)=a(x^2-6x+13)(x-r).$$Moreover, $a=1$, because we are given that $p(x)$ is monic.

Substituting $x=0$, we have $p(0)=-13r$, but we also know that $p(0)=-52$; therefore, $r=4$. Hence we have $$\begin{array}{rl}p(x)& =(x^2-6x+13)(x-4)\\ & =\boxed{x^3-10x^2+37x-52}.\end{array}$$Solution #2 (essentially the same as #1, but written in terms of using Vieta's formulas)

Since $p(x)$ has real coefficients and has $3-2i$ as a root, it also has the complex conjugate, $3+2i$, as a root. The sum and product of these two roots, respectively, are $6$ and $3^2-(2i{)}^2=13$. Thus, the monic quadratic that has these two roots is $x^2-6x+13$.

By the Factor Theorem, we know that $x^2-6x+13$ divides $p(x)$. Since $p(x)$ is cubic, it has one more root $r$. Because $p(0)$ equals the constant term, and because $p(x)$ is monic, Vieta's formulas tell us that $(3-2i)(3+2i)r=(-1{)}^3(-52)=52$. Thus $r=4$, and $$\begin{array}{rl}p(x)& =(x^2-6x+13)(x-4)\\ & =\boxed{x^3-10x^2+37x-52}.\end{array}$$