1988 United States presidential election in Utah
The 1988 United States presidential election in Utah took place on November 8, 1988. All fifty states and the District of Columbia, were part of the 1988 United States presidential election. State voters chose five electors to the Electoral College, which selected the president and vice president.
Utah was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.
Utah weighed in for this election as 26% more Republican than the national average.Partisan background
The presidential election of 1988 was a very partisan election for Utah, with over 98 percent of the electorate in voting for either the Republican or Democratic parties, though several other parties appeared on the ballot. Every county in Utah voted in majority for the Republican candidate, except for Carbon County, which voted primarily for Dukakis.Republican victory
Bush won the election in Utah with a 34-point sweep-out landslide – his strongest victory in the nation. The election results in Utah are somewhat reflective of a nationwide political reconsolidation of base for the Republican Party, which took place through the 1980s. Through the passage of some very controversial economic programs, spearheaded by then President Ronald Reagan, the mid-to-late 1980s saw a period of economic growth and stability. The hallmark for Reaganomics was, in part, the wide-scale deregulation of corporate interests, and tax cuts for the wealthy.
Dukakis ran his campaign on a socially liberal platform, and advocated for higher economic regulation and environmental protection. Bush, alternatively, ran on a campaign of continuing the social and economic policies of former President Reagan – which gained him much support with social conservatives and people living in rural areas.Results
Results by county
