<s>[INST] Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$. 
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.

My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is zero at minus zero point something and minus one point something.
So, even at local minima, the cubic is positive. It means it would cross x-axis only once.
At $x=-2$, cubic is positive and at $-3$, it is negative.
It means the only root is minus two point something.
Is there any other way to solve this question? Something that doesn't involve calculator? Or maybe something that doesn't involve calculus?
 [/INST] Since $ \ x^2 + x + 1 \ = \ \left(x + \frac12 \right)^2 + \frac34 \ > \ 0 \ $ for all real numbers, we might ask under what circumstances $ \ (x^2+x+1)^2+2 \ > \ (x^2+x+1)(x^2-2x-6) \ \ , \ $ since the product on the right side can plainly take on negative values.  Dividing out a factor of $ \ (x^2+x+1) \ $ produces
$$    x^2  \ + \ x \ + \ 1  \ + \ \frac{2}{x^2+x+1} \ \ > \ \   x^2 \ - \ 2x \ - \ 6  \ \ \Rightarrow \ \ \frac{2}{x^2+x+1} \ \ > \ \  - 3x \ - \ 7  \ \ . $$
We will not need to solve for the exact interval over which this inequality holds.  The left side of the inequality remains positive for all real numbers, while the linear function on the right side is positive only for $ \ x \ < \ -\frac73 \ \ . $  Hence, this inequality is certainly true for $ \ x \ \ge \ -\frac73 \ \ . \ $  This tells us that the intersection of the curves representing the functions on the two sides of the original equation occurs at a value $ \ c \ < \ -\frac73 \ \ $  (in fact, the intersection is at $ \ c \  \approx \ -2.477 \ ) \ , \ $  so there can be no roots of the original equation in the interval $ \ (-2 \ , \ 4) \ \ . \ $
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