<s>[INST] How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$.  I have 
$(2x^2-1x)(6x-3)$ 
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring? 
 [/INST] So we can actually generalize this. Say we have the polynomial 
$$p(x)=ax^2+bx+c$$
Fact:
$$p(x)=\bigg(ax+\frac{b-\sqrt{b^2-4ac}}{2}\bigg)\bigg(ax+\frac{b+\sqrt{b^2-4ac}}{2}\bigg)$$

Proof:
Let's assume the existence of three real numbers $r_1$, $r_2$, and $e_1$ such that 
$$ax^2+bx+c=e_1(x-r_1)(x-r_2)$$
If we expand the product on the right hand side and then compare coefficients,
$$ax^2+bx+c=e_1x^2-e_1(r_1+r_2)x+e_1r_1r_2$$
we get a system of equations
$$e_1=a\\-e_1(r_1+r_2)=b\\e_1r_1r_2=c$$
Evidently, we get $e_1=a$ for free. So we update our system of equations:
$$r_1+r_2=-\frac{b}{a}\\r_1r_2=\frac{c}{a}$$
We can solve each equation for $r_2$:
$$r_2=-\frac{b}{a}-r_1\\r_2=\frac{c}{ar_1}$$
So we can set the two equations equal to each-other:
$$-\frac{b}{a}-r_1=\frac{c}{ar_1}$$
$$r_1+\frac{c}{ar_1}=-\frac{b}{a}$$
multiplying both sides by $ar_1$,
$$ar_1^2+br_1=-c$$
Then we add $\frac{b^2}{4a}$ to both sides:
$$ar_1^2+br_1+\frac{b^2}{4a}=\frac{b^2}{4a}-c$$
Then we note that 
$$a(r_1+b/2a)^2=ar_1^2+br_1+\frac{b^2}{4a}$$
So we plug it in:
$$a(r_1+b/2a)^2=\frac{b^2}{4a}-c$$
$$a(r_1+b/2a)^2=\frac{b^2-4ac}{4a}$$
$$(r_1+b/2a)^2=\frac{b^2-4ac}{4a^2}$$
$$r_1+b/2a=\sqrt{\frac{b^2-4ac}{4a^2}}$$
$$r_1+b/2a=\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$$
$$r_1+b/2a=\frac{\sqrt{b^2-4ac}}{2a}$$
$$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
And since we know that 
$$r_2=-\frac{b}{a}-r_1$$
We know that 
$$r_2=-\frac{b}{a}-\bigg(\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)$$
$$r_2=-\frac{2b}{2a}+\frac{b-\sqrt{b^2-4ac}}{2a}$$
$$r_2=\frac{b-2b-\sqrt{b^2-4ac}}{2a}$$
$$r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
And by definition,
$$ax^2+bx+c=a\bigg(x+\frac{b-\sqrt{b^2-4ac}}{2a}\bigg)\bigg(x+\frac{b+\sqrt{b^2-4ac}}{2a}\bigg)$$
$$ax^2+bx+c=\bigg(ax+\frac{b-\sqrt{b^2-4ac}}{2}\bigg)\bigg(ax+\frac{b+\sqrt{b^2-4ac}}{2}\bigg)$$
And with that our proof is complete :)
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