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1 solution
( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) = ( x − 1 ) ( x − 4 ) ( x − 2 ) ( x − 3 ) = ( x 2 − 5 x + 4 ) ( x 2 − 5 x + 6 ) = ( x 2 − 5 x + 5 − 1 ) ( x 2 − 5 x + 5 + 1 ) = ( x 2 − 5 x + 5 ) 2 − 1 2
Since the term ( x 2 − 5 x + 5 ) 2 ≥ 0 and ( x 2 − 5 x + 5 ) 2 = 0 when x = 2 5 ± 5 , then the minimum value of ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) = 0 − 1 = − 1 .