### Instruction: Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$.

Determine $p(x)$ (in expanded form).### Response: I know that a monic cubic polynomial has the form $p(x)=x^3+ax^2+bx+c$, where $a$, $b$, and $c$ are real coefficients.
I also know that if a polynomial has complex roots, then they must occur in conjugate pairs.
That means if $p(3-2i)=0$, then $p(3+2i)=0$ as well.
So, $p(x)$ must have three roots: $3-2i$, $3+2i$, and some real number $r$.
This means that $p(x)$ can be factored as $p(x)=(x-3+2i)(x-3-2i)(x-r)$, where $r$ is unknown.
I can use the given information that $p(0)=-52$ to find $r$.
Plugging in $x=0$ into the factored form, I get $p(0)=(-3+2i)(-3-2i)(-r)=-52$.
Simplifying, I get $(9+4)(-r)=-52$, or $r=\frac{52}{13}=4$.
So, $p(x)=(x-3+2i)(x-3-2i)(x-4)$.
To get the expanded form, I need to multiply out the factors.
I can start by multiplying the first two factors, which are conjugates of each other.
I can use the difference of squares formula to simplify: $(x-3+2i)(x-3-2i)=(x-3)^2-(2i)^2=x^2-6x+9-(-4)=x^2-6x+13$.
Now, I need to multiply this by $(x-4)$.
I can use the distributive property to do this: $(x^2-6x+13)(x-4)=x^3-6x^2+13x-4x^2+24x-52=x^3-10x^2+37x-52$.