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 $a \ne 1$?**
 A: 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 $x$.
A: 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. 
A: Algebra students learn that the quadratic formula says
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
When is this true? It's helpful to remember that the quadratic formula comes from the following statement, which is true for any complex numbers $a$, $b$, $c$, and $x$ (as long as $a\ne0$):
$$\text{If }ax^2+bx+c=0,\quad\text{then }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
So the quadratic formula is good for getting the roots of a quadratic polynomial. Now if your goal is to factor a quadratic polynomial, then it's helpful to know the following:
$$\text{If }p,\,q\text{ are the roots of }ax^2+bx+c,\quad\text{then }ax^2+bx+c=a(x-p)(x-q).$$
Hence, if you want to factor $2x^2+3x+1$, then you could note that
$$2x^2+3x+1=2(x-p)(x-q)$$
where $p$, $q$ are the roots of $2x^2+3x+1$. Since the quadratic formula says that the roots of $2x^2+3x+1$ are $-\frac{1}{2}$ and $-1$, we have that
$$2x^2+3x+1=2\left(x+\frac{1}{2}\right)\left(x+1\right).$$
A: 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: $ax^2+bx+c = g(x) => a\cdot(x-r_1)(x-r_2) = g(x)$
