Intuitive understanding of factoring quadratic equations When factoring a second degree equation $ax^2 + bx + c$ you find the roots then take $a(x - \text{root})(x - \text{root})$. I am wondering why this works. Sorry if poorly phrased question.

We can show the solutions to $ax^2+bx+c=0$ is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$  by completing the square and solving for $x$. So that means the factored form of $ax^2+bx+c$ can be written as $a(x-\frac{-b + \sqrt{b^2-4ac}}{2a})(x-\frac{-b - \sqrt{b^2-4ac}}{2a})$. Expanding this gives: $a(x+\frac{b - \sqrt{b^2-4ac}}{2a})(x+\frac{b + \sqrt{b^2-4ac}}{2a})=a(x^2+(\frac{b- \sqrt{b^2-4ac}}{2a}+\frac{b+\sqrt{b^2-4ac}}{2a})x+\frac{b^2-(b^2-4ac)}{4a^2})=(x^2+\frac{b}{a}x+\frac{4ac}{4a^2})=ax^2+bx+c$ 
