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A materials testing technician is modeling the flex point yy (in inches) of a new polymer using the equation 16y2=32y3+2y{}16y^2 = 32y^3 + 2y. After rearranging the equation to 32y3βˆ’16y2+2y=0{}32y^3 - 16y^2 + 2y = 0 and completely factoring it to 2y(4yβˆ’1)2=0{}2y(4y - 1)^2 = 0, the technician applies the Zero Product Property. The two distinct solutions for the flex point yy are 0{}0 and ____.

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Updated 2026-05-25

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