Question

f(x)=x^3-4x^2-3x +12 are√3and -√3 find the third zero

Answer 1

$let \: the \: zeroes \: be \: \alpha = \sqrt{3} \: and \: \beta = - \sqrt{3}$

$(x - \alpha )(x - \beta ) \\ = (x - \sqrt{3} )(x + \sqrt{3}) \\ = {x}^{2} - { (\sqrt{3}) }^{2} \\ = {x}^{2} - 3$

One factor is (x² – 3)

Another factor is
$\frac{ {x}^{3} - {4}^{2} - 3x + 12 }{ {x}^{3} - 3} \\ = x - 4$
For zero,
x – 4 = 0
=> x = 4

Therefore, the third zero is 4.



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