Brewster's angle is only a special case and not as crucial as people might think in this case. To keep things short:
Mirror: Conventional mirrors have a metallic surface (e.g. aluminum or silver) where the mobile surface charge carriers cause a reflection of the light. Polarization usually does not matter enough here to cause a significant polarization of the reflected light. Hence, it can only be partially blocked by a polarization filter. You can look at Graphs for s- and p-polarization on this page (click on "Graphs" for plots) to verify that s- and p-pol don't make too much of a difference here.
Floor: Dielectric (non-conducting) materials also reflect a part of the light according to the Fresnel equations (This page actually shows a figure related to your question). For large angles of incidence (let's say roughly larger than $40°$) the reflection properties are usually completely different for s- and p-pol. P-pol light is then reflected poorly while a part of the s-pol light still gets reflected. Brewster's angle is an extreme case of this ($0\%$ reflection of p-pol light), but generally the p-pol reflection is much weaker for steep angles. This causes a strong polarization of the reflection.