Question 1172300
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(1)  x^2 - 11x + 28 = factoring = (x-7)*(x-4).    (1)



(2)  You are given that 

         x^2 - 11x + 28 is a factor of x^4 + k(x^3) - 67(x^2) + 394x - 504.


     It means that each factor of  (1),  (x-7) and (x-4)  are factors of the polynomial  

         p(x) = x^4 + k(x^3) - 67(x^2) + 394x - 504.      (2) 



(3)  In turn, due to the remainder theorem, it means that  the value of x= 4 (as well as the value of x= 7) 

     is the root of the polynomial (2).  It gives you an equation for k

         p(4) = 0 = 4^4 + k*4^3 - 67*4^2 + 394*4 - 504.


     It is the same as

                256 + 64k = 0,


     which implies  k = -256/64 = -4.


     So,  k = -4.



(4)  The sum of the roots of the equation 

         x^4 + k(x^3) - 67(x^2) + 394x - 504 = 0      (3)


     is equal to the coefficient value at x^3  with the opposite sign  (the Vieta's theorem).


     It means that the sum of the roots of the equation (3) is equal to -k = 4.


<U>ANSWER</U>.  The sum of the roots of the equation  x^4 + k(x^3) - 67(x^2) + 394x - 504 = 0  is equal to 4.
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Solved.



&nbsp;&nbsp;&nbsp;&nbsp; * * * The miracle is that I answered the problem's question WITHOUT solving equation (3). &nbsp;&nbsp;&nbsp;&nbsp;* (!) * (!) *