### Instruction: Find a monic quadratic polynomial $P(x)$ with integer coefficients such that
\[P(\sqrt{13} - 3) = 0.\](A polynomial is monic if its leading coefficient is 1.).### Response: I want to find a quadratic polynomial of the form $P(x) = x^2 + bx + c$ where $b$ and $c$ are integers, and $P(\sqrt{13} - 3) = 0$.
If I plug in $\sqrt{13} - 3$ for $x$, I get
\[(\sqrt{13} - 3)^2 + b(\sqrt{13} - 3) + c = 0.\]
Expanding and simplifying, I get
\[13 - 6\sqrt{13} + 9 + b\sqrt{13} - 3b + c = 0.\]
Now I can group the terms with and without $\sqrt{13}$:
\[(22 - 3b + c) + (b - 6)\sqrt{13} = 0.\]
For this equation to hold, both coefficients must be zero, so I have a system of equations:
\[\begin{cases}
22 - 3b + c = 0\\
b - 6 = 0
\end{cases}\]
Solving for $b$, I get $b = 6$.
Plugging this into the first equation, I get
\[22 - 18 + c = 0,\]
which implies $c = -4$.
Therefore, the quadratic polynomial I want is
\[P(x) = x^2 + 6x - 4.\]