Question

Please solve the following questions with steps

Answer 1

$hey \: mate \: here \: is \: your \: answer...$
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$p(x) = 2 {x}^{3} + {x}^{2} - 6x - 3$

$\: \: \: \: \: \: \: \: \: \: \: \: x = - \sqrt{3}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \sqrt{3} \\ x + \sqrt{3} = 0...(1) \: \: \: \: \: \: x - \sqrt{3 } = 0...(2)$

$on \: multiplying \: equation \: (1) \: and \: (2)$

$\: \: \: \: \: (x + \sqrt{3} )(x - \sqrt{3} ) = 0 \\ {x}^{2} - \sqrt{3} x + \sqrt{3} x - 3 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} - 3 = 0$
$on \: dividing \: p(x) \: by \: {x}^{2} - 3$

$after \: dividing \: we \: get \:$

$2x + 1 = 0 \\ \: \: \: \: \: \: \: 2x = - 1 \\ \: \: \: \: \: \: \: \: \: x = \frac{ - 1}{2}$
$the \: other \: zero \: of \: the \: polynomial \: is \: \frac{ - 1}{2}$

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$hope \: it \: helps.....$