Question 1181497
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If -2 is a root of ((z^3) - (8(z^2)) + (9z) + 58 = 0, what are the other two roots?
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In fact, -2 is a root of the given polynomial, and you can check it in one line, 
substituting the value of -2 into the polynomial

    {{{(-2)^3}}} - {{{8*(-2)^2}}} + {{{9*(-2)}}} + 58 = -8 - 32 - 18 + 58 = 0.



Therefore, the given polynomial is divisible by (z-2) without a remainer (according to the Remainder theorem)

and you can divide it using standard long division procedure 


    {{{(z^3 - 8z^2 + 9z + 58)/(z-2)}}} = z^2 - 10z + 29.


Solve this quadratic equation using the Quadratic Formula


    {{{z[1,2]}}} = {{{(10 +- sqrt(10^2 - 4*29))/2}}} = {{{(10 +- sqrt(-16))/2}}} = {{{(10 +- 4i)/2}}} = 5 +- 2i.


<U>ANSWER</U>.  Two other roots of the given equation are  (5+2i)  and  (5-2i).
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Solved.