====== Factoring with an X ======
This is the method I was taught to factor with, and I still use it to this day. I can't find the proper name of this((diamond?? x-box??)) so X-Factoring is what I'll call it.

===== Tutorial Values =====
This tutorial assumes the standard form quadratic $ax^2 + bx + c$. For the steps below, I'll use the example quadratic $8x^2 - 72x + 162$.

Here's a diagram if the X if you get confused.

{{:math:diagram.png?200|}}
===== Step 0 // GCF Factor =====
The first step before even starting to factor is to look for a GCF in all three terms, it will simplify the problem immensely if there is an $a$ value((you can still factor even without doing this, it'll just be harder)). In the case of our example, $8x^2 - 72x + 162$ can be simplified down to $2(4x^2 - 36x + 81)$. This may save you from factoring all together((say, if you are solving for roots and you are able to GCF factor in a way that lets you skip directly to using zero product property)).

===== Step 1 // Drawing the X  =====
The first real step in factoring is to draw an X. In the top section, place your $c$ value in, and for the bottom, your $b$ value. For our example (after GCF factoring), the $c$ value of 81 goes on top, and the $b$ value of 36 goes below.

{{:math:step_0.png?200|}}

===== Step 2 // Multipliers and a =====
Now, for the two multipliers, they will be the factors of $a$. If there is no $a$ value, this factoring method is extremely simple as they will both be $1$, thus it's practically plug 'n chug. However, if there is an $a$ value, and especially if it is a composite number, it can get complicated quickly. For this example I chose one with an $a$ value to entice the brain a bit. If it is a prime number, the modifiers are easy as it **has** to be $1$ and the $a$ value. Note that the order of the modifiers does **not** matter((if you think about it, the x and y values would just swap places if they were flipped)). For composite numbers, you may have to guess and check for possible modifier/$x$/$y$ values, and here a factor tree for both the $a$ and $c$ values are extremely helpful. From anecdotal experience, try using the factors of $a$ that are closest together first (2 and 2 first before 4 and 1) as they tend to be correct more often than not. __2 and 2 are the correct multipliers for the example.__

{{:math:step_1.png?200|}}

===== Step 3 // Finding x and y =====
Now this is the bulk of the mental gymnastics, you must find two values, that when multiplied equal $c$, and add to equal $b$ after being multiplied by the multiplier below the number. Again, a factor tree is helpful if $c$ is a composite number. Likewise, if it its a prime number, the two values must be $1$ and the $c$ value. There isn't really a systematic method, you just have to quickly guess and check values until you find two that fit both requirements. You get better at this as you use this method more. __For the example values of 9 for both $x$ and $y$ satisfy the requirements.__

{{:math:step_2.png?200|}}

===== Step 4 // Double Check =====
To double check, ensure that $xy = c$ and $y_{\text{modifier}} \times x + x_{\text{modifier}} \times y = b$. Or, in simpler terms, make sure $x$ and $y$ multiply to $c$ and add to $b$, accounting for multipliers below the numbers if present.
  * $\boxed{9 \times 9 = 81} \checkmark$
  * $\boxed{2 \times 9 + 2 \times 9 = 36} \checkmark$

===== Step 5 // Writing the Solution =====
To write the solution, put the x, y, and modifier values in the form $(x_{\text{modifier}}t + x)(y_{\text{modifier}}t + y)$((using $t$ as the $x$ input variable to not confuse that $x$ **with the $x$ from the X-Factoring diagram**)). In other words, the x and y value use the multiplier on the opposite side of them, hence why they are labeled as such in the [[math:x_factoring#tutorial_values|Tutorial Values]] section. The answer for the GCF'd example is $(2x + 9)(2x + 9)$((which can simplify to $(2x + 9)^2$)). Thus, after including the GCF, $8x^2 - 72x + 162$ factors to $\boxed{2(2x + 9)(2x + 9)}$ which is our answer.

{{:math:step_3.png?200|}}((FIXME whoops the image has a ^2 in both factors, ignore that!))