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.
 A: $$4x^2+2x+1$$
Finding roots by quadratic rule as follows 
$$x=\frac{-2\pm\sqrt{2^2-4(4)(1)}}{2(4)}$$ $$=\frac{-2\pm2i\sqrt{3}}{8}$$
$$=-\frac{1}{4}\pm\frac{i\sqrt 3}{4}$$
$$x=-\frac{1}{4}+\frac{i\sqrt 3}{4}\ \vee\ x=-\frac{1}{4}-\frac{i\sqrt 3}{4}$$
Edit:

Now, we have the factors as follows $$4x^2+2x+1=4\left(x+\frac{1}{4}-\frac{i\sqrt 3}{4}\right)\left(x+\frac{1}{4}+\frac{i\sqrt 3}{4}\right)$$

A: Given $$\displaystyle 4x^2+2x+1 = \underbrace{(2x)^2+1^2+\left(\frac{1}{2}\right)^2}+1-\underbrace{\left(\frac{1}{2}\right)^2}$$
So we get $$\displaystyle \left(2x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}i}{2}\right)^2 = \left(2x+\frac{1}{2}+\frac{\sqrt{3}i}{2}\right)\cdot \left(2x+\frac{1}{2}-\frac{\sqrt{3}i}{2}\right)$$
A: 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 
$\frac{-1- \sqrt {-3}}{4}=\frac{1}{-1+\sqrt {-3}}$)
A: 
$$4x^2+2x+1=3x^2+(x+1)^2=(\sqrt3x+i(x+1))(\sqrt3x-i(x+1))$$

A: You could always check if they are the same by multiplying out the course notes's solution.  Start by writing the proposed solution
$$(1-(-1+\sqrt{-3})x)(1-(-1-\sqrt{-3})x)$$
as
$$( (1+x) - \sqrt{-3}x ) ( (1 + x) + \sqrt{-3}x )$$
(this just involves some regrouping).  Now this has the form $(A - B)(A + B)$, so we can immediately rewrite it as
$$(1+x)^2 - (-3)x^2$$
which, on expanding, gives
$$1 + 2x + x^2 + 3x^2$$
which is indeed the original problem.
Having said that, it is also the case that the way the Notes have the solution written is not standard, and I am not sure why they prefer to do it this way. 
