<s>[INST] Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have 
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots: 
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-write the original quadratic as: 
$$(x-1)\left(x-\tfrac{1}{2}\right)$$
However, if you actually multiply this out then you get: 
$x^2-\frac{3}{2}x + \frac{1}{2}$ which is not the same as  $2x^2-3x+1$. 
So why the discrepancy? Aren't you supposed to get the original equation? 
 [/INST] How can $x=\frac{1}{2}$ and $x=1$ both be solutions to $x^2 - \tfrac{3}{2}x+\tfrac{1}{2} = 0$ and $2x^2-3x+1=0$?
The simple fact is that $2x^2-3x+1 \equiv 2\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so
$2x^2-3x+1 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.
By a similar argument: $4x^2-6x+2 \equiv 4\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so
$4x^2-6x+2 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.
In general, for any non-zero number $k$, we have
$ax^2+bx+c=0$ if, and only if, $(ka)x^2+(kb)x+(kc)=0$. Why not try to prove this?!
Even more generally, for any non-zero number $k$: $\mathrm{f}(x)=0$ if, and only if, $k\cdot \mathrm{f}(x)=0$.
</s>