clock menu more-arrow no yes

Filed under:

Seahawks sit all alone atop NFC standings following Packers’ blowout loss to Buccaneers

Green Bay Packers v Tampa Bay Buccaneers Photo by Mike Ehrmann/Getty Images

The Seattle Seahawks didn’t play this weekend but they were winners in terms of the standings.

A stifling Tampa Bay Buccaneers defense held Aaron Rodgers to one of the worst performances of his career, completing just 16/35 for 160 yards and two interceptions, one of which was returned for a pick-six by Jamel Dean. An early 10-0 lead for the Green Bay Packers turned into a 38-10 blowout loss, meaning the Packers are no longer unbeaten and the Seahawks are the last undefeated team in the NFC.

It has been yet another season thus far of very close, nail-biting wins for the Seahawks even though their schedule has consisted of a pair of 1-5 teams, the 2-3 New England Patriots, 2-3 Dallas Cowboys, and 3-3 Miami Dolphins, but you can’t control who you play and a win is a win.

Six weeks into the season, being first place in the NFC West this doesn’t really mean a whole lot. But any chance the Seahawks can benefit from playoff rivals losing, you take them. There are three one-loss teams now in the NFC: The Packers, Chicago Bears, and Los Angeles Rams. If the San Francisco 49ers could pull off an upset against the Los Angeles Rams, the Seahawks would be two ahead in the loss column in the NFC West on their nearest rivals. That would be the ideal weekend since a double loss cannot happen to the Rams and 49ers.