Question

18y^3 - 2y factorize
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Answer 1

$18 {y}^{3} - 2y = 0\\ 2y(9 {y}^{2} - 1) = 0 \\ 2y = 0 \\ y = 0 \\ \\ 9 {y}^{2} - 1 = 0 \\ {y}^{2} = \frac{1}{9} \\ y = \frac{1}{3} \\ factors \: are \: 2y \: and \: (9 {y}^{2} - 1) \\ and \: zeroes \: are \: 0 \: and \: \frac{1}{3}$