I know that if a quadratic polynomial with real coefficients has a complex root, then its complex conjugate must also be a root. That means that if $-2 - 3i$ is a root, then $-2 + 3i$ must also be a root. So the polynomial must have the form $p(x) = a(x - (-2 - 3i))(x - (-2 + 3i))$ for some real constant $a$. I can simplify this expression by multiplying out the factors and using the fact that $i^2 = -1$. I get $p(x) = a(x + 2 + 3i)(x + 2 - 3i) = a(x^2 + 4x + 13)$. Now I need to find the value of $a$ that makes the coefficient of $x$ equal to $-4$. That means I need to solve $4a = -4$, which gives me $a = -1$. So the quadratic polynomial is $p(x) = -x^2 - 4x - 13$. 
Find the quadratic polynomial, with real coefficients, which has $-2 - 3i$ as a root, and where the coefficient of $x$ is $-4$.