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 $a \ne 1$?**

Your error is in forgetting about the leading coefficient. If a monic polynomial $p(x) $ has roots $r_1,\ldots, r_n$, with multiplicity, then
$$p(x) =(x-r_1) \cdots(x-r_n) $$
If it is not monic but instead has a leading coefficient of $a$, then
$$p(x) =a(x-r_1) \cdots(x-r_n)$$
Note that for this to work you need to include all of the roots. For polynomials with real coefficients, some of these roots may be complex. 
