Q. \[ x^2 + x + 1 \]

We are asked to work with the expression \(x^2 + x + 1\). Since the problem statement only shows this expression, there is no additional instruction (such as factoring, finding roots, simplifying, expanding, etc.). So the most helpful step-by-step response is to analyze common typical tasks for a quadratic like this: factoring, completing the square, and finding its discriminant/roots.

Step 1: Identify the type of expression

The expression \(x^2 + x + 1\) is a quadratic polynomial in \(x\), with coefficients:

\[
a = 1,\quad b = 1,\quad c = 1
\]

Step 2: Check whether it factors nicely over the integers

To factor a quadratic \(x^2 + x + 1\) over the integers, we look for two integers \(m\) and \(n\) such that:

\[
x^2 + x + 1 = (x + m)(x + n)
\]
This would require:

\[
m + n = 1,\quad mn = 1
\]

The integer pairs that satisfy \(mn = 1\) are \((m,n) = (1,1)\) or \((-1,-1)\).

If \((m,n) = (1,1)\), then \(m + n = 2\), not \(1\). If \((m,n) = (-1,-1)\), then \(m + n = -2\), not \(1\).

Therefore, it does not factor over the integers.

Step 3: Complete the square

Start with:

\[
x^2 + x + 1
\]

Group the first two terms and factor out the coefficient needed for completing the square:

\[
x^2 + x + 1 = \left(x^2 + x\right) + 1
\]

Take the coefficient of \(x\), which is \(1\), and halve it:

\[
\frac{1}{2}
\]

Square that value to get what we must add and subtract:

\[
\left(\frac{1}{2}\right)^2 = \frac{1}{4}
\]

Now rewrite \(x^2 + x + 1\) by adding and subtracting \(\frac{1}{4}\):

\[
x^2 + x + 1 = \left(x^2 + x + \frac{1}{4}\right) + 1 – \frac{1}{4}
\]

The expression \(x^2 + x + \frac{1}{4}\) is a perfect square:

\[
x^2 + x + \frac{1}{4} = \left(x + \frac{1}{2}\right)^2
\]

Also simplify \(1 – \frac{1}{4}\):

\[
1 – \frac{1}{4} = \frac{3}{4}
\]

So the completed-square form is:

\[
x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}
\]

Step 4: Interpret what this tells us (often used to check roots)

Because \(\left(x + \frac{1}{2}\right)^2 \ge 0\) for all real \(x\), we have:

\[
x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4}
\]

This means \(x^2 + x + 1\) is always positive for real \(x\), so it has no real zeros.

Final result (useful simplified form)

\[
x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}
\]