The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
Algebra Precalculus – How to Re-write Completing the Square for $x^2+x+1$
algebra-precalculuscompleting-the-squarepolynomialsquadratics
Related Solutions
Your factoring solution is correct, and you can see it's correct by plugging in the solutions into the original formula. To complete the square, the first thing we want to do is divide both sides by the coefficient on the $x^2$ term, which is $6$ for this problem. Doing this gives $x^2 + 4x = 0$. Adding $4$ to both sides completes the square:
\begin{align} &x^2 + 4x = 0 \\ &x^2 + 4x + 4 = 4 \\ &(x+2)^2 = 4 \end{align}
To finish up, take a square root of both sides (and don't forget about the $\pm$ when doing this).
P.S. the quadratic formula $x = \displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ that gives the solutions to an arbitrary $ax^2 + bx + c = 0$ can be derived with this completing the square method.
"I just don't understand why we can complete the square."
The best way to see why it works, is to see a pictorial representation of what is going on, see:
The $x^2$ term is a square with both sides of length $x$. The $bx$ term is a rectangle with one side of length $x$ and one side of length $b$. When you divide by two you are splitting the $bx$ rectangle in two parts with still a side of length $x$ and the other side is now $\frac b2$.
When you place both rectangles as shown on the picture there is only a small square with sides of length $\frac b2$ missing. This small square has area $\frac {b^2}{4}$ and it is this part that you need to "complete the square".
If you have a term $a\ne 1$ in $ax^2+bx+c$ then you have to first "get it off" to get a square with both sides of length $x$. In that case the rectangle you get still has a side of length $x$ but its other side is now $\frac ba$ and this makes the area of the tiny missing square equal to $\frac {b^2}{4a}$.
Best Answer
Remember the formula for the square of a binomial: $$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is, $$x^2 + x + \cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=\frac{1}{2}$.
But if you have $(x+\frac{1}{2})^2$, you get $x^2 + x + \frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$\frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $\frac{1}{4}$. So: \begin{align*} x^2 + x + 2 &= (x^2 + x + \cdots) + 1\\ &=\left( x^2 + 2\left(\frac{1}{2}\right)x + \cdots \right) + 1 &&\mbox{figuring out what $c$ is}\\ &= \left(x^2 + 2\left(\frac{1}{2}\right)x + \left(\frac{1}{2}\right)^2\right) -\frac{1}{4} + 1 &&\mbox{completing the square}\\ &= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}. \end{align*}