SOLUTION: For what value(s) of k will x^2+4x+4 be a factor of: f(x)= x^4−18x^2+kx+44

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Question 1172123: For what value(s) of k will x^2+4x+4 be a factor of: f(x)= x^4−18x^2+kx+44

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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Notice that  x%5E2+%2B+4x+%2B+4 = %28x%2B2%29%5E2. 


Therefore, if the trinomial x%5E2+%2B+4x+%2B+4  divides  the polynomial  f(x)= x%5E4-18x%5E2%2Bkx%2B44  with no remainder,

it implies that the binomial  (x+2)  divides  this polynomial   with no remainder.



In turn, due to the Remainder theorem, it means that x= -2 is the root of the polynomial f(x).



So, we substitute  x= -2 into the polynomial f(x) and equate it to zero

    f(-2) = %28-2%29%5E4-18%28-2%29%5E2%2Bk%28-2%29%2B44 = -12 - 2k = 0.


It gives us a linear equation for k, which leads to 

    2k = -12,

     k = -12/2 = -6.



At this point, we obtain the following result

    +-----------------------------------------------------------------+
    |  if the trinomial  x^2+4x+4  divides  the polynomial            |  
    |                                                                 |
    |  f(x)= x^4-18x^2+kx+44  with no remainder, then k= -6.          |
    +-----------------------------------------------------------------+



But the direct check shows that 

    +-----------------------------------------------------------------+
    |    the trinomial  x^2+4x+4  DOES NOT  divide the polynomial     |
    |                                                                 |
    |            f(x)= x^4-18x^2-6x+44    with no remainder           |
    +-----------------------------------------------------------------+


Hence, the value of "k" satisfying the given condition DOES NOT EXIST.


ANSWER.  The value of "k" satisfying the given condition DOES NOT EXIST.

Solved, answered, explained and completed.