Question

Find the zeroes of y^2 - 2y -8

Answer 1

Explanation:

${y}^{2} - 2y - 8 = 0$

By Splitting the Middle term, we get

${y}^{2} - 4y + 2y - 8 = 0$

$y(y - 4) + 2(y - 4)$

$(y + 2)(y - 4)$

therefore,

$y + 2 = 0 \: \: \: \: and \: \: \: \: y - 4 = 0$

$y = - 2 \: \: \: and \: \: \: y = 4$

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