Find the monic quadratic polynomial, in $x,$ with real coefficients, which has $-2 - i \sqrt{5}$ as a root.
To find a monic quadratic polynomial with real coefficients, I need to use the fact that complex roots always come in conjugate pairs. That means if $-2 - i \sqrt{5}$ is a root, then so is $-2 + i \sqrt{5}.$ Therefore, the polynomial must be of the form $(x - (-2 - i \sqrt{5}))(x - (-2 + i \sqrt{5})) = 0.$ To simplify this expression, I need to multiply out the two binomials using the distributive property and the rules for multiplying complex numbers. I get $(x + 2 + i \sqrt{5})(x + 2 - i \sqrt{5}) = x^2 + 4x + 4 - (i \sqrt{5})^2 = x^2 + 4x + 4 + 5 = x^2 + 4x + 9.$ 