If the squared difference of the zeroes of the quadratic polynomial f(x) = x^2 + px + 45 is equal to 144, find the value of p.

Let zeroes be $\alpha$ and eta. We have \alpha + eta = -p and \alphaeta = 45. Given (\alpha - eta)^2 = 144, which expands to (\alpha + eta)^2 - 4\alphaeta = 144. Substituting the values: $(-p)^2 - 4(45) = 144 \implies p^2 - 180 = 144 \implies p^2 = 324$, so $p = \pm 18$.