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What are the zeros of the function g(x)=x^3-3x^2-4x+12

User Kevin He
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5 votes

Answer:

_2, 2, 3

Explanation:


g(x) = {x}^(3) - 3 {x}^(2) - 4x + 12 \\ = {x}^(2) (x - 3) - 4(x - 3) \\ = (x - 3)( {x}^(2) - 4) \\ = (x - 3)(x - 2)(x + 2) \\ plug \: g(x) = 0 \\ \therefore \: (x - 3)(x - 2)(x + 2) = 0 \\ (x - 3) = 0 \: or \: (x - 2) = 0 \: \\ or \: (x + 2) = 0 \\ \therefore \:x = 3 \: or \: x = 2 \: or \: x = - 2 \\ \therefore \:x = \{ - 2, \: \: 2, \: \: 3 \}

Hence, - 2, 2 & 3 are the zeros of the given function.

User Sergey Chikuyonok
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