Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot  2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1]$$
$$x_1 = -1/2,~~~x_2 = -1$$
therefore:
$$2x^2 + 3x + 1 = (x + 1/2)(x+1)$$
Now I check it:
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x \cdot x + x \cdot 1 + \frac{1}{2}\cdot x + \frac{1}{2}\cdot 1$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + \frac{3}{2}x + \frac{1}{2}$$
but:
$$2x^2 + 3x + 1 \ne x^2 + \frac{3}{2}x + \frac{1}{2}$$
Why does't quadratic formula work when $a \ne 1$?
however, I can pull out the 1/2.
$$2x^2 + 3x + 1 \ne \frac{1}{2}[2x^2 + 3x + 1]$$
I feel that this is something I must have missed in grade school...
Does this mean you need to multiple by "a" if $a \ne 1$?**

You are not "solving" $$2x^2+3x+1$$
but $$
2x^2+3x+1  \mathbf{=0}.
$$
For the specific values of $x$ you found, it is indeed true that
$$
2x^2+3x+1 = x^2+\frac{3}{2}x+\frac{1}{2}=0 \quad \text{ where }x\in \left\{- \frac{1}{2}, -1\right\}
$$
Since we solved for when this expression is equal to zero, we are free to multiply everything by any factor to obtain a different true equation:
$$
2x^2+3x+1 = A\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=0
$$
for any real number $A$ when $x$ is one of $-1/2$ or $-1$.
It is not true that in general
$$
2x^2+3x+1 = x^2+\frac{3}{2}x+\frac{1}{2}
$$ for any $x$.
