It is true that if you quickly recognise a factor, that it will be faster than using the quadratic formula, but unfortunately, that is actually a difficult task for a classical computer at this point (I say 'classical' in comparison to a super Turing machine like an artificial neural network or a quantum computer that doesn't rely on classical systems). However, the Quadratic Formula helps to close that gap for classical machines *and* humans by providing an easy method to factor a quadratics.

That being said, I typically use the completing the square method (I think that is what it is called) if I cannot recognise the factors quickly enough. **EDIT:** Completing the square can be used to easily derive the quadratic formula— just plug it in directly for ax^{2}+bx+c and you get precisely the quadratic formula.

My guess, though, is that the reason the teachers didn't teach it is that they wanted you to do it for yourself. If your curriculum runs like mine did way back when I first learned it, they will next teach you about completing the square (another, longer method solving quadratic equations) and then use that to develop the quadratic formula. Knowing how it works is really important to understanding math, not just that it works.

Another method, as follows, does not require a program, but requires a little guessing. But, it is fast and easy, and accurate for integers:

1) turn on calculator

2) press the Y= button in the top left corner (in the ti 83 and 84, at least)

3) enter your equation

4) hit graph button

5) check for where the graph crosses the x-axis (the horizontal one). In your case, there likely will be two points (X

An example: if the graph appears to cross the x-axis at (7,0) and (-4,0), as is the case for your given problem, then the factors appear to be (X-7) and (X+4). Note the sign changes.

Also, you can check these answers by, in graph mode, hitting the trace button and entering the values of 7 and -4 (checking one at a time). If you are right, then the corresponding y-values will equal 0.

The reason this works is that, when the whole equation equals 0, at least one factor must equal zero. If necessary, I can try to give a more in depth explanation.

P.S. - this also works for finding the real roots (X_{subscripts}) of other degree polynomials, such as aX^{3} + bX^{2} + cX + d.

\begin{align} x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{align}

So for example, using x^{2}-3x-28, solving for x we get:

\begin{align} x=\frac{--3\pm \sqrt{9-4*-28}}{2}=\frac{3\pm \sqrt{121}}{2}=\frac{3\pm 11}{2} \end{align}

So x=7 or x=-4. These tell us where the quadratic is 0, so in equation form, we have (x-7)(x+4). If you plug in -4 or 7 to this, you get 0 and this is the factored form of your quadratic :)

]]>Otherwise, the easiest way is just to use the quadratic formula. Your example equation would produce 7 and -4, so the factors must be (x-7) and (x+4).

]]>Now, I have this summer assignment due for my upcoming freshmen year, and this factoring section is killing me. The problems are basic but super long and tedious, and us programmers are lazy. The problems are like this:

**Simplify**

x²-3x-28

a²+bx+c

I was taught to find the products of a and c and find the factor pair which added up to b. Then you would do this:

x²-3x-28

x²+4x-7x-28

x(x+4)-7(x+4)

(x-7)(x+4)

I guess most of you know this. Anyways, it takes a million years to find that factor pair when the numbers get huge and ugly. So I wanted a program that would do it for me. I had my draft input the product of a and c and then b. I tried using a For( loop but the List issues are what make me stuck. Here's the code (so far)

` ``PROGRAM:FACTOR :ClrList L₁,L₂ :ClrHome :Input "AC: ",X :Input "B: ",Y :For(A,1,X :X/A→B :If fPart(B)=1 :Then`

I don't know what to do after that Then statement. I want to place all the pairs which are potential in a list of some sort, but I don't know how to add variables into a list (I mean that any number of variables could be placed into the list, ranging from 1 to infinity) and I don't know how to extract a certain pair from a list. I'm not sure if this is possible…

One of my numerous thoughts was to put another For loop inside this For loop, and once "Potential Pair A" is found, it uses this second For loop to find its placement inside List 1 and 2. Once all potential pairs are found, it uses a third For loop (outside the first and second) to run through the lists and find the pair which has the sum of b. Then it extracts it and displays it. The issue is that I don't know if this is possible and I know for a fact it could take a super fast time running through a potentially massive pair of lists.

Another theory is to look on the internet for some already-thought-about equation or theorem that finds the solution to the entire factoring problem. Problem is that if it existed my teachers would have already shown it to us. The other is that I am feeling lazy, leading to the entire thought about this program.

So if anyone has their own copy of this programming issue or has any ideas on how to accomplish this, please replay ASAP!!! I'm not a math algorithm guru, so I need your help, people!

L1K3AWH15P3R

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