Solution: Note $ x^2 - 2x + 1 = (x - 1)^2 $. Use polynomial division or remainder theorem for repeated roots. Let $ f(x) = x^5 - 3x^3 + 2x - 1 $. The remainder $ R(x) $ has degree < 2, so $ R(x) = ax + b $. Since $ (x - 1)^2 $ divides $ f(x) - R(x) $, we have $ f(1) = R(1) $ and $ f'(1) = R'(1) $. Compute $ f(1) = 1 - 3 + 2 - 1 = -1 $. $ f'(x) = 5x^4 - 9x^2 + 2 $, so $ f'(1) = 5 - 9 + 2 = -2 $. $ R(x) = ax + b $, so $ R(1) = a + b = -1 $, $ R'(x) = a $, so $ a = -2 $. Then $ -2 + b = -1 $ → $ b = 1 $. Thus, remainder is $ -2x + 1 $. Final answer: $ oxed-2x + 1 $.Question: A plant biologist is studying a genetic trait that appears in every 12th plant in a rows of crops planted in a 120-plant grid. If the trait is expressed only when the plant’s position number is relatively prime to 12, how many plants in the first 120 positions exhibit the trait? - Crosslake