Find the zeroes of polynomial $(x^{3} - 5 x^{2} - 16 x + 80)$ . If its two zeroes are equal in magnitude but opposite in sign.

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Solution

Zeroes of polynomial $x^{3} - 5 x^{2} - 16 x + 80$ if two zeroes are equal in magnitude and apposite in sign.
Let the zero be $^{'} a^{'}$
Other zero will be $- a$
Let unit zero be $^{'} b^{'}$
Sum of three zeroes $= 5 \Rightarrow a + (- a) + b = 5$
$\Rightarrow$ $b = 5$
Product of three zeroes $= - 80 \Rightarrow a (- a) (b) = - 80$
substituting in above equation $b = 5$
$\Rightarrow - a^{2} b = - 80 \Rightarrow - a^{2} (5) = - 80 \Rightarrow a = \pm 4$
$\Rightarrow$ other two roots are $- 4, 4$
Hence, the zeroes are $4, - 4, 5.$
Hence, the answer is $4, - 4, 5.$

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