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 $a \ne 1$?**

On the quadratic formula:
$$ax^2 +bx+c=0$$
$$r_i =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$a\cdot(x-r_1)(x-r_2)$$
First, the "$a\cdot$" term in the factored result "$a\cdot(x-r_1)(x-r_2)$" is an often overlooked term when using the quadratic formula to factor a second order polynomial, because usually a=1, or the "$a\cdot$" term gets divided out by 0 on the RHS.  So people forget that they need to add this "$a\cdot$" to the factorized result, but its really there, even if it doesn't matter because it will get canceled by the zero on the RHS.
Second, quadratic formula really applies to only  the LHS of the expression $ax^2 +bx+c=0$.   Its just that algebra textbooks usually only care about factoring the LHS when the RHS is zero.  However, its perfectly ok to use the quadratic formula to factor the LHS when the right hand side is not zero. Example: $ax^2+bx+c = g(x) => a\cdot(x-r_1)(x-r_2) = g(x)$
