Question

Find the zeros of the polynomial x^3-5x^2-16x+80 if two of its zeroes are equal in magnitude but opposite in sign

Answer 1

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Answer 2

Appropriate Question:

$\sf{find \: the \: zeroes \: of \: the \: polynomial \: p(x) = x^3 -5x^2 - 16x + 80, if \: it s \: two \: zeroes \: are \: equal \: in \: magnitude \: but \: opposite \: in \: sign.}$

Solution:

$\bf{we \: have \: p(x) = {x}^{3} - {5x}^{2} - 16x + 80 }$

$\sf{let \: one \: zero \: be \: \alpha .then, \: another \: zero( \beta ) = - \alpha }$

Now,

$\bf{ \alpha + \beta + \gamma = \frac{ - b}{a} } \\ : \implies \bf{ \alpha + ( - \alpha ) \gamma = \frac{ -( - 5)}{1} } \\ : \implies \bf{ \gamma = 5 } \: \ \: \: \: \: \: \fbox {(1)}$

Also,

$: \implies \bf{ \alpha \beta \gamma = \frac{ - d}{a} } \\ : \implies \bf{ \alpha ( - \alpha ) \gamma = - 80} \\ : \implies \bf{ { \alpha }^{2} \gamma = - 80 } \\ : \implies \bf{ { - \alpha }^{2} \times 5 = - 80 } \\ : \implies \bf{ { - \alpha }^{2} = \frac{ - 80}{5} } \\ : \implies \bf{ { - \alpha }^{2} = - 16 } \\ : \implies \bf{ { \alpha }^{2} = 16 } \\ : \implies \bf{ \alpha = \pm4}$

$\pink{\mathbb{CASE \: 1} }\leadsto \mathcal{ WHEN \: \alpha = + 4} \\ \mathcal{THEN, \: \beta = - 4 } \\ \mathfrak {( \orange{\because} \red{ \alpha \: and \: \beta are \: equal \: in \: magnitude \: but \: opposite \: in \: sign})} \\ \mathcal{So, \: ZEROS \: ARE : 4,-4,5.}$

$\pink{\mathbb{CASE \: 2} }\leadsto \mathcal{ WHEN \: \alpha = - 4. } \\ \mathcal{THEN, \: \beta = 4 \: \: AND \: \: \gamma = 5} \\ \mathcal{So, \: ZEROS \: ARE : -4,4 \: AND \: 5.}$