Yes, exactly! The proof is based on a contradiction: if a polynomial of degree n does not have any complex solutions, then it can be factored into n linear factors with real coefficients. But we know that every real polynomial can be factored into linear factors, so this means that the original polynomial must have had complex solutions. By assuming the opposite, we reach a contradiction, which means that the original assumption must be false, and therefore every polynomial must have at least one complex solution.
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