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Patriots Signing TE Austin Hooper To One-Year, $4.25M Deal
Jamie Sabau-USA TODAY Sports

According to Ian Rapoport, the Patriots are expected to sign TE Austin Hooper to a one-year deal worth up to $4.25 million. 

Hooper, 29, was drafted in the third round by the Falcons out of Stanford in 2016. He played out the final year of a four-year, $3.165 million rookie contract and made a base salary of $720,000 for the 2019 season. 

Hooper was testing the market as an unrestricted free agent when he signed a four-year, $42 million deal with the Browns. He was set to make a base salary of $9.5 million in the final two years of the deal when the Browns released him in 2022. 

The Titans would later sign Hooper to a one-year, $6 million contract. He signed a one-year deal with the Raiders last offseason. 

In 2023, Hooper appeared in all 17 games for the Raiders and recorded 25 receptions for 234 yards (9.4 YPC) and no touchdowns. 

This article first appeared on NFLTradeRumors.co and was syndicated with permission.

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