Find the monic quadratic polynomial, in $x,$ with real coefficients, which has $1 - i$ as a root.
I know that a monic quadratic polynomial has the form $x^2 + bx + c,$ where $b$ and $c$ are real coefficients. I also know that if a polynomial with real coefficients has a complex root, then its conjugate must be another root. Therefore, the other root of the polynomial must be $1 + i.$ To find $b$ and $c,$ I can use the fact that the sum of the roots is $-b$ and the product of the roots is $c.$ The sum of the roots is $(1 - i) + (1 + i) = 2,$ so $b = -2.$ The product of the roots is $(1 - i)(1 + i) = 1 - i^2 = 2,$ so $c = 2.$ Therefore, the monic quadratic polynomial is $x^2 - 2x + 2.$ 