<s>[INST] 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 $a \ne 1$?**
 [/INST] 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
$$2x^2+3x+1=2\left(x+\frac{1}{2}\right)\left(x+1\right).$$
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