Math, asked by wowgvurav, 4 months ago

find the zeroes of the p(x)=(x-3)(x+2)

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Answered by umitbarman1111

4

Answer:

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Answered by singhdipanshu2707200

1

Answer:

1)p

$p(x) = {x}^{2} - 1x - 6 \\ \: = (x - 3) = 0 \\ x = 3 \\ (x + 2) = 0 \\ x = - 2 \\ so \: zero \: is \: 3 \: \: or - 2$

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