Question 1052839
if you use synthetic division and use the possible values of x to be equal to plus or minus 1 or 2, you will find that the graph crosses the x-axis at x = 2


the equation becomes (x-2) * (x^2 + x + 1) = 0


if you try to factor x^2 + x + 1, you will find that the roots are complex.


this should mean that the graph only crosses the x-axis at x = 2.


when x < 2, the graph is negative.


when x > 2, the graph is positive.


the solution should be that the equation is > 0 when x > 2.


here's the graph of the equation.


<img src = "http://theo.x10hosting.com/2016/101406.jpg" alt="$$$" </>


the concept involved is as follows:


let a = (x-2)


let b = (x^2 + x + 1)


a * b = 0 if a = 0 or if b = 0 or if a and b are both equal to 0.


a = (x-2) = 0 when x = 2


b = (x^2 + x + 1) is never equal to 0, because the value of x when b = 0 is not real.


in other words, the graph of x^2 + x + 1 never crosses the x-axis because its roots are not real.


therefore, the only possibility for a * b = 0 is when a = 0 which is when (x-2) is equal to 0 which occurs when x = 2.


the graph confirms the logic.


the logic appears to be sound.


i did a couple of tests to see if this was true and it appears that it is.


either way, your solution is that the equation is > 0 when x > 2.