Question

what should be added to (z²+24z+120)so that the resulting polynomial is exactly divisible by (z+12)​

Answer 1

Given p(x) = $\rm z^{2} + 24z + 120$

g(x) = $\rm z+12$

By long division method, we get.

$\rm z + 12 \sqrt{ {z}^{2} + 24z + 120 } \\ {z}^{2} ± 12z \\ -------- \\ \: \: \: \: \: \: \: \: \: \: \: 12z + 120 \\ \: \: \: \: \: \: \: \: \: \: 12z ± 144 \\ - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -(+) 24 \\ - - - - - - - -$

Therefore, 24 should be added to $\rm z{2} + 24z + 120$ so that it exactly get's divisible by $\rm (z+12)$

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Note :

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