Quadratic formula question: Missing multiplying factor of A? I have a very simple problem which must have a simple answer and I was wondering if anyone can point out my error.
I have the following quadratic equation to factor:
$2x^2+5x+1$
Which is of the form:
$Ax^2+Bx+C$
All I want to now do is factor this into the form:
$(x+\alpha)(x+\beta)$
So using the quadratic formula
$x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$
I get
$x = \frac{-5 \pm \sqrt{17}}{4}$
So I would think then that
$(x - \frac{-5 + \sqrt{17}}{4})(x - \frac{-5 - \sqrt{17}}{4}) = 2x^2+5x+1 $
But it doesn't seem to work. Any x that I choose results in the answer being off by a factor of A (in this case A=2).
What silly thing am I missing?
 A: When you use the quadratic formula to find the roots of a quadratic, you search for solutions $x$ to the quadratic equation $$Ax^2 + Bx + C = 0$$
Note that this is equivalent to the quadratic equation $(A\neq 0)$ $$x^2 + \frac BA\cdot x+ \frac CA = 0$$ 
Both the quadratics $Ax^2 + Bx + C$ and $x^2 + \frac BAx +\frac CA$ share the same roots.
A: We can obtain the correct factorization by completing the square.
\begin{align*}
2x^2 + 5x + 1 & = 2\left(x^2 + \frac{5}{2}x\right) + 1\\
              & = 2\left(x^2 + \frac{5}{2}x + \frac{25}{16}\right) - \frac{25}{8} + 1\\
              & = 2\left(x + \frac{5}{4}\right)^2 - \frac{17}{8}\\
              & = 2\left[\left(x + \frac{5}{4}\right)^2 - \frac{17}{16}\right]\\
              & = 2\left[x + \frac{5}{4} + \frac{\sqrt{17}}{4}\right]\left[x + \frac{5}{4} - \frac{\sqrt{17}}{4}\right]\\
              & = 2\left[x + \frac{5 + \sqrt{17}}{4}\right]\left[x - \frac{5 + \sqrt{17}}{4}\right]
\end{align*}
However, when we solve the equation $2x^2 + 5x + 1 = 0$ by completing the square, we first transform it into the monic equation 
$$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$
whose roots are the same as those of the original equation.  When we derive the Quadratic Formula, we complete the square on the equation
$$ax^2 + bx + c = 0$$
where $a \neq 0$.  The first step in the derivation is to divide $ax^2 + bx + c = 0$ by $a$ to transform it into the monic equation
$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
whose roots are equivalent to those of the equation $ax^2 + bx + c = 0$.
If we factor the monic equation, we obtain
$$\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) = 0$$
The sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.
To obtain the original equation from the monic equation, we must multiply the monic equation by $a$.  Thus, the factorization of the original equation is 
\begin{align*}
ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)\\
              & = a\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)
\end{align*}
What you found by using the sum and product of the roots was the factorization of the monic equation
$$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$
To obtain the original equation, you must multiply your factorization by $a = 2$.
A: $Ax^2+Bx+C$ factorize as : $A(x-\alpha)(x-\beta)$ wiht $\alpha + \beta= \dfrac{-B}{A}$ and $\alpha \beta=\dfrac{C}{A}$.
obviously if $A \ne 0$, and  $\alpha$ and $\beta$ are roots of the the two equations:
$$
A(x-\alpha)(x-\beta)=0 \iff Ax^2+Bx+C=0
$$
and
$$
(x-\alpha)(x-\beta)=0 \iff x^2+\dfrac{B}{A}x+\dfrac{C}{A}=0
$$
