1912 United States presidential election in Iowa
The 1912 United States presidential election in Iowa took place on November 5, 1912, as part of the 1912 United States presidential election. Voters chose thirteen representatives, or electors, to the Electoral College, who voted for president and vice president.
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Elections in Iowa | ||||||||||
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Iowa was won by Princeton University President Woodrow Wilson (D–Virginia), running with governor of Indiana Thomas R. Marshall, with 37.64% of the popular vote, against the 26th president of the United States Theodore Roosevelt (P–New York), running with governor of California Hiram Johnson, with 32.87% of the popular vote and the 27th president of the United States William Howard Taft (R–Ohio), running with Columbia University President Nicholas Murray Butler, with 24.33% of the popular vote.[1]
Wilson was the first Democrat to win Iowa's electoral votes since Franklin Pierce in 1852. No Democrat would obtain a majority until Franklin Roosevelt in 1932.
Results
Party | Candidate | Votes | % | |
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Democratic | Woodrow Wilson | 185,325 | 37.64% | |
Progressive | Theodore Roosevelt | 161,819 | 32.87% | |
Republican | William Howard Taft | 119,805 | 24.33% | |
Socialist | Eugene V. Debs | 16,967 | 3.45% | |
Prohibition | Stuart Hamblen | 8,440 | 1.71% | |
Total votes | 492,356 | 100% |