Easy quadratic problem

Knowing that $\frac{2x^2 + 2x + 3}{x^2 + x + 1} = 2 + \frac{1}{x^2+x+1}$ , we require (x^2 + x + 1)|1 for $x \in \mathbb{R}$ . Hence:

$x^2 + x + 1 = \pm1$ .

If it's equal to 1, then $x^2 + x = 0 \Rightarrow x = 0,-1$ . If it's equal to -1, then $x^2 + x + 2 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{7}i}{2}$ and we do not have real roots! Therefore the least integral value of $p$ such that $2 + \frac{1}{x^2+x+1} \le p$ holds is $\boxed{3}.$

$\frac{2[x^{2} +x +1] +1}{[x^{2} +x +1]}$ , Let the polynomial be $X$

$\frac{2X +1}{X}$ = $2+\frac{1}{X}$

$2+\frac{1}{X} \leq p$

Lowest possible answer to this inequality is 2 when X approach $\infty$

Since we need integral value that is defined, taking limits is not an option because 2 is never reached and since the polynomial doesn't take negative values the only possible answer here is 3.