Boston Red Sox today signed first baseman Steve Pearce to a one-year contract through the 2019 season. Sox President of Baseball Operations Dave Dombrowski made the announcement.
Prior to being named Most Valuable Player of the World Series, Pearce, 35, split the 2018 regular season between the Toronto Blue Jays and Red Sox, combining to hit .284 (61-for-215) with an .890 OPS, 11 home runs, and 42 RBI in 76 games. He made 28 starts at first base, 19 at designated hitter, seven in left field, and two in right field. Acquired from Toronto with cash considerations on June 28 in exchange for minor league infielder Santiago Espinal, Pearce appeared in 50 games with Boston, batting .279 (38-for-136) with a .394 on-base percentage. For the Red Sox, he hit five home runs against the New York Yankees, including three on August 2.
“We’re thrilled to have Steve back with us for another year as we think he’s a great fit for our club,” said Dombrowski. “Obviously, we all saw what kind of impact he can have on the field, especially with the Postseason that he had. He also provides good depth and balance from the right side for us.”
Against left-handed pitchers in 2018, Pearce batted .304 (31-for-102) with a .400 on-base percentage and .559 slugging percentage. Following the regular season, he started 11 of the Red Sox’ 14 postseason games, all at first base. He reached base via hit or walk in each of his 12 playoff games, batting .289 (11-for-38) with four home runs, 11 RBI, a 1.083 OPS, and nine walks against eight strikeouts. After homering in Game 4 of the Fall Classic at Dodger Stadium and adding two home runs in Game 5, Pearce was named the Willie Mays World Series Most Valuable Player. In the Fall Classic, each of his four hits went for extra bases and he drew four walks without striking out, posting a .500 on-base percentage with eight RBI against the Dodgers.