How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (-1 + \sqrt{-3})x)$ and $(1 - (-1 - \sqrt{-3})x)$  
Course Notes (pg. 30) 
Are these two expressions equivalent, or am I simply not factoring correctly? Thanks.

The solution is $\frac{-1 \pm \sqrt {-3}}{4}$ (suppose we know what we mean here by $\sqrt{-3}$).
Then the expansion would be
$\begin{align*}
&4x^2+2x+1=4\left(x-\frac{-1+ \sqrt {-3}}{4}\right)\left(x-\frac{-1- \sqrt {-3}}{4}\right)\\
& = 4\left(x-\frac{1}{-1- \sqrt {-3}}\right)\left(x-\frac{1}{-1+ \sqrt {-3}}\right)\\
&=[-1+(-1- \sqrt {-3})x][-1+(-1+ \sqrt {-3})x] \\
&=[1-(-1- \sqrt {-3})x][1-(-1+ \sqrt {-3})x]
\end{align*}$
 as written in the lecture note. 
Note that here we've used the fact that $(-1+ \sqrt {-3})(-1- \sqrt {-3})=4$.
(Thus $\frac{-1+ \sqrt {-3}}{4}=\frac{1}{-1- \sqrt {-3}}$
And 
$\frac{-1- \sqrt {-3}}{4}=\frac{1}{-1+\sqrt {-3}}$)
