Question 1048804
set the equation equal to 0.
you get (x-5) * (x^2 + x + 1) = 0
this is true if (x-5) = 0 and/or if (x^2 + x + 1) = 0


x-5 = 0 when x = 5.


this means that x-5 is negative when x < 5 and x-5 is positive when x > 5.


x^2 + x + 1 never equals 0.
in fact, it is always positive.


since x^2 + x + 1 = 0 is a quadratic equation in standard form, then it has a max/min point at x = -b/2a.


since a = 1 and b = 1 and c = 1, that max/min point will be at x = -1/2.


when x = -1/2, the max/min point will be at y = (-1/2)^2 - 1/2 + 1 which will be at y =  1/4 - 1/2 + 1 which becomes y = 3/4.


since the coefficient of the x^2 term is positive, then x = -1/2 is a min point and the equation opens up.


that says that all values of y = x^2 + x + 1 are positive regardless of the value of x.


not only are they all positive, but they are all greater than or equal to 3/4.


since x^2 + x + 1 is always positive and since x-5 is positive when x > 5 and negative when x < 5, then:


(x-5) * (x^2 + x + 1) is positive when x > 5 because positive times positive always gives a positiv
e result.

(x-5) * (x^2 + x + 1) is negative when x < 5 because positive times negative always give a negative result.


your solution is that (x-5) * (x^2 + x + 1) > 0 when x > 5.


the following graph confirms that this is true.


<img src = "http://theo.x10hosting.com/2016/091901.jpg" alt="$$$" </>


here's the graph of x-5 by itself.


<img src = "http://theo.x10hosting.com/2016/091902.jpg" alt="$$$" </>


here's the graph of x^2 + x + 1 by itself.


<img src = "http://theo.x10hosting.com/2016/091903.jpg" alt="$$$" </>