SOLUTION: Given that x^2 - 11x + 28 is a factor of x^4 + k(x^3) - 67(x^2) + 394x - 504, evaluate the sum of the four roots of the equation x^4 + k(x^3) - 67(x^2) + 394x

SOLUTION: Given that x^2 - 11x + 28 is a factor of x^4 + k(x^3) - 67(x^2) + 394x - 504, evaluate the sum of the four roots of the equation x^4 + k(x^3) - 67(x^2) + 394x - 504 = 0

Question 1172300: Given that x^2 - 11x + 28 is a factor of x^4 + k(x^3) - 67(x^2) + 394x - 504, evaluate the sum of the four roots of the equation x^4 + k(x^3) - 67(x^2) + 394x - 504 = 0
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

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

Solved.

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

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Given that x^2 - 11x + 28 is a factor of x^4 + k(x^3) - 67(x^2) + 394x - 504, evaluate the sum of the four roots of the equation x^4 + k(x^3) - 67(x^2) + 394x - 504 = 0

k = - 4
Roots of
Sum of the roots of
You're welcome, TUTOR @ IKLEYN! I wasn't CHECKING you though!
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