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? 
 A: Since you are applying quadratic formula to factorize, I am assuming that you made a typo in the degree of x in the first term. It should be $2x^2$ and not $2x^3$ 
Also,
$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is the correct factorization of the trinomial.
$f(x)=0$ does have roots $x_1$ and $x_2$, but the multiplicand '$a$' $=2$ is missing in your reconstruction of the trinomial as a product of its factors.
A: The quadratic formula gives the solution to the equation
$$ax^2+bx+c=0.$$
But the formula is derived by completing the square, and to do so the coefficient of $x^2$ must be 1. Therefore, the first step is to divide by $a$ (it is assumed without loss of generality that $a>0$). This gives
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$$
Now notice that 
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=\left(x-\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\right)\left(x-\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\right).$$
Long story short, the quadratic formula gives the solutions of $ax^2+bx+c=0$ by finding the solutions to the $equivalent$ equation $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$
A: Your linear factor is $(2x-1)$, i.e.,
$$
2x^2-3x+1=(x-1)(2x-1)\neq (x-1)(x-1/2).
$$
A: 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, $k\cdot \mathrm{f}(x)=0$.
