Question

Q1 The zeroes of x2 + 2x - 8 are​

Answer 1

Answer:

$x ^{2} + 2x - 8 = 0 \\ \\ = > x ^{2} + (4 - 2)x - 8 = 0 \\ \\ = > x ^{2} + 4x - 2x - 8 = 0 \\ \\ = > x(x + 4) - 2(x + 4) = 0 \\ \\ = > (x + 4)(x - 2) = 0 \\ \\ either \: x + 4 = 0 \\ \\ = > x = - 4 \\ \\ or \: x - 2 = 0 \\ \\ = > x = 2$

⭐ Zeroes aRe - 4 and 2 .

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