# Is there a valid, “shortcut” method to convert a quadratic from vertex form to factored form?

Recently, I had a couple of students give me an interesting way of converting from vertex form to factored form that I've never seen before. One told me he got it from his science teacher, and another one got it from his dad. Neither of them could give me much information beyond that.

The equation I asked them to change to factored form was $f(x) = (x-3)^2 - 1$.

They first wrote it as the factors $(x-3-1)(x-3+1)$, and then simplified to $(x-4)(x-2)$.

I've tried to search around for a source for this method and turned up nothing. As far as I can tell, they just got lucky with the fact that the vertex happened to have a y-value of $-1$, and a leading coefficient of $1$, meaning the $x$-intercepts just so happened to be one unit to the left and right of $x=3$.

However, if there is a magical way to change from vertex to factored form in about $4$ seconds flat, I'd like to know about it. I could be missing some rules for using this trick.

So is there a valid, shortcut-type method for converting a quadratic equation in vertex form directly to factored form?

• This is just $a^2-b^2=(a+b)(a-b)$ with $a=(x-3)$, $b=1$. – user856 Nov 30 '17 at 1:17

This is called the difference of two squares: \begin{align}f(x) &= (x-a)^2 - b \\&= \left(x-a + \sqrt{b}\right)\left(x-a - \sqrt{b}\right)\\&=\left(x-(a-\sqrt{b})\right)\left(x-(a + \sqrt{b})\right)\end{align} You can check this easily: $(c-d)(c+d) = c^2 - dc + cd - d^2 = c^2 - d^2$, and apply this with $c=(x-a)$ and $d = \sqrt{b}$.