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
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-> 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
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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(52797) (Show Source):
(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). * (!) * (!) *
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!