Question

zeros of the polynomial Z cube​

Answer 1

SOLUTION

TO DETERMINE

The zeros of the polynomial

$\sf{ {z}^{3} - 8 }$

EVALUATION

Here the given polynomial is

$\sf{ {z}^{3} - 8 }$

Now we find the zeroes as below

$\sf{ {z}^{3} - 8 = 0}$

$\sf{ \implies \: {z}^{3} - {2}^{3} = 0}$

$\sf{ \implies \:(z - 2)( {z}^{2} + 2z + 4) = 0 }$

Now z - 2 = 0 gives z = 2

Now

$\sf{ ( {z}^{2} + 2z + 4) = 0 }$

$\displaystyle\sf{ \implies z = \frac{ - 2 \pm \: \sqrt{ {2}^{2} - 4 \times 1 \times 4} }{2 \times 1} }$

$\displaystyle\sf{ \implies z = \frac{ - 2 \pm \: \sqrt{ 4 - 16} }{2 \times 1} }$

$\displaystyle\sf{ \implies z = \frac{ - 2 \pm \: \sqrt{ - 12} }{2 } }$

$\displaystyle\sf{ \implies z = \frac{ - 2 \pm \: 2 \sqrt{ - 3} }{2 } }$

$\displaystyle\sf{ \implies z = - 1 \pm \: \sqrt{ - 3} }$

Hence the required zeroes are

$\sf{2 \: ,\: - 1 + \sqrt{3} \: ,\: - 1 - \sqrt{3} }$

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