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Assuming you are trying to solve "(x - 1)(x + 2)(x - 3) = 0", when any number is multiplied by zero, the result is always zero, regardless of the other number(s). Thus each term in parentheses can be made zero in turn to find the solutions:

x - 1 = 0 → x = 1, giving (1 -1)(1 + 2)(1 - 3) = 0 × 3 × -2 = 0

etc.

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Q: Can we use principle zero products to solve (x-1)(x 2)(x-3)?
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