The expression \( x^2 + x + 1 \) is a quadratic polynomial. Here are a few things you might want to know about it:

1. **Factoring**: This particular quadratic does not factor nicely over the real numbers because its discriminant (\( \Delta = b^2 - 4ac \)) is negative. For \( x^2 + x + 1 \), the discriminant is:
\[
\Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3
\]
Since the discriminant is negative, the polynomial has no real roots and cannot be factored into real linear factors.

2. **Roots**: The roots can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). For \( x^2 + x + 1 \):
\[
x = \frac{-1 \pm \sqrt{-3}}{2 \cdot 1} = \frac{-1 \pm \sqrt{3}i}{2}
\]
So the roots are:
\[
x = \frac{-1 + \sqrt{3}i}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{3}i}{2}
\]

3. **Vertex**: The vertex of the parabola represented by the quadratic equation \( y = x^2 + x + 1 \) can be found using the vertex formula \( x = -\frac{b}{2a} \). For \( x^2 + x + 1 \):
\[
x = -\frac{1}{2 \cdot 1} = -\frac{1}{2}
\]
Plugging this back into the equation to find the y-coordinate:
\[
y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}
\]
So the vertex is at \( \left( -\frac{1}{2}, \frac{3}{4} \right) \).

4. **Graph**: The graph of \( y = x^2 + x + 1 \) is a parabola that opens upwards. It has no x-intercepts (since it has no real roots), and its vertex is at \( \left( -\frac{1}{2}, \frac{3}{4} \right) \).

If you have any specific questions or need further details about this quadratic polynomial, feel free to ask!