Factoring quadratics - why does this method work? Given a quadratic like this:
$2x^2 + 7x + 3$
I need to find two numbers $a$ and $b$ that have these properties:
$a + b = 7$
$a \cdot b = 2 \cdot 3$
Which in this case is 1 and 6. 
So I can rewrite the equation as:
$  2x^2 + 1x + 6x + 3$
$= x(2x + 1) + 3(2x + 1)$
$= (2x + 1) (x + 3)$
I wanted to know why this method works. I found a proof in a khan academy video that goes like this, based on the multiplication of two binomials:
$(fx + g) (hx + j)$
$= fhx^2 + ghx + jfx + gj$
$= fhx^2 + x(gh + jf) + gj$
Then let's assume that $gh = a$ and $jf = b$ this would be the proof for finding the middle coefficients. I understand this part.
The video then tells me that therefore $a \cdot b = gh \cdot jf$ which can be rewritten as $a \cdot b = fh \cdot gj$. 
I don't understand this part - because for me it doesn't really prove that I need to find a number that multiplies the first coefficient with the constant (the last number) in the equation. Is there maybe another way to explain that?
 A: Let's assume $ax^2 + bx + c = (rx + s)(tx + u)$.  Then 
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
              & = rx(tx + u) + s(tx + u)\\
              & = rtx^2 + rux + stx + su\\
              & = rtx^2 + (ru + st)x + su
\end{align*}
is an algebraic identity that holds for every real number $x$.  In particular, it holds if $x = 0$, $x = 1$, and $x = -1$.  If $x = 0$, we obtain 
$$c = su \tag{1}$$
If $x = 1$, we obtain 
$$a + b + c = rt + ru + st + su \tag{2}$$
Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain 
$$a + b = rt + ru + st \tag{3}$$
If $x = -1$, we obtain
$$a - b + c = rt - ru - st + su \tag{4}$$
Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain
$$a - b = rt - ru - st \tag{5}$$
Adding equations 3 and 5 yields
$$2a = 2rt \tag{6}$$
Dividing both sides of equation 6 by 2 yields
$$a = rt \tag{7}$$
Since $a = rt$, we can cancel $a$ from the left hand side of equation 3 and $rt$ from the right hand side of equation 3 to obtain
$$b = ru + st \tag{8}$$
Furthermore, observe that 
$$ac = (rt)(su) = (ru)(st) \tag{9}$$
Hence, we can factor a polynomial with rational coefficients with respect to the rational numbers if there exist two numbers ($ru$ and $st$) with product $ac$ and sum $b$.
A: This method uses four parameters ($f,g,h,j$) when the initial trinomial only has three, making things a little confusing. I am showing a simpler version.
From $$px^2+qx+r$$
you can factor out the coefficient $p$, giving
$$px^2+qx+r=p\left(x^2+\frac qpx+\frac rp\right).$$
Now it should be obvious that the factorization is of the form
$$px^2+qx+r=p(x+a)(x+b)=p(x^2+(a+b)x+ab).$$
With the given example
$$2\left(x^2+\frac72x+\frac32\right),$$ and you need to solve
$$\begin{cases}a+b=\dfrac72,\\a\,b=\dfrac32\end{cases}.$$
Then, notice that
$$(a+b)^2-4ab=(a-b)^2.$$
Using this relation,
$$a-b=\pm\sqrt{\frac{7^2}{2^2}-4\cdot\frac32}=\pm\frac52.$$
Then combining with the known sum,
$$a,b=\frac12,3$$
and 
$$2x^2+7x+3=2\left(x+\frac12\right)\left(x+3\right)=(2x+1)(x+3).$$
A: Suppose that we have 
$$ax^2+bx+c$$
One strategy is to restate this as a quadratic a leading coefficient equal to unity.
$$\begin{align}
ax^2+bx+c&=\frac{a}{a}(ax^2+bx+c)\\
&=\frac{1}{a}(a^2x^2+bax+ac)\\
&=\frac{1}{a}((ax)^2+b(ax)+ac)\\
&=\frac{1}{a}(t^2+bt+ac)\\
\end{align}$$
Where $t=ax$
Now, we're just looking for two numbers that add together to make $b$ and multiply together to make $ac$. Afterwards, just replace the $t$s with $ax$s.
