# Understanding Algebra: Why do we factor equations?

What's algebra about? When learning about variables (x, y, z), they seem to "hide" a number:

\$\$x + 3 = 5\$\$

This is a companion discussion topic for the original entry at http://betterexplained.com/articles/understanding-algebra-why-do-we-factor-equations/

I went back to college almost a year ago and despite hating math when I was younger, I am loving it now. I’m nearly finished with Calculus 1, and while I got all A’s in algebra, I never really understood why one would set a polynomial equation to zero. Obviously I knew that the math worked out this way, but I’ve never had a professor explain it as you just did (or even try to explain it at all for that matter). I really enjoyed your error analogy. It instantly clicked for me, like a light was switched on. I won’t be able to stop thinking about it for a little while.

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This so nice

Hi Robert, happy the analogy clicked! I’m having a blast going back and trying to find metaphors for all the mathematical truths I memorized but didn’t internalize. Appreciate the comment.

@ Robbie

Are you sure you wrote the question properly? There are two “/” signs in there, making it a bit confusing as to whether you’re dividing the denominator or numerator.

Im a junior high student and the teacher is teaching factoring, Im good at the simple factoring but Im stuck in this question:
(2 x^2+x-6)/(x^3-3 x^2+2)(x^2+4 x-5)/(4 x^2-6 x)
I would really appreciate it if you could do it step by step and explain it slowly
Thx
Best Regards
Robbie

@Chris: Thanks!

@hdhoang: Great find, thank you. Now to figure out how it works :).

@gjing: Nice, glad it worked. I don’t have much experience with PDEs but I think that analogy could be extended. “Factoring” is just a process used to set up something which can be zero’d… there could be other ways.

@mark: Hah, I wish I had an mental error function! (Hello stock market picks). Exactly, finding the roots of the derivative can give us better insights. And again, we factor the derivative to find when “the changes are zero” (i.e., any required component of the entire change process becomes zero).

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Reverse engineering
\$\$(x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6\$\$

But can’t figure out the intuition on the reverse the splitting of the x?

Ehhh \$\$2 * 3 = 6\$\$ nope still no light

Must be that it’s 25 year ago that i was in school but you lost me at
Onto The Factoring
How do you get from \$\$x^2 + x-6\$\$ to \$\$(x+3)(x-2)\$\$

I like the breaking the system though

Hi Franc, great question! For polynomials with only a squared term, a common factoring trick is to see what numbers could multiply together to get 6.

As you reverse-engineered, the last two numbers (x - a)(x - b) need to multiply to get 6:

\$\$a * b = 6\$\$

In this case, we have choices 1 & 6, 2 & 3, 3 & 2, or 6 & 1. At this point it’s trial and error to find two factors with a difference of 1 (i.e., leaving us with a single “x”). In this case it’s 2 and 3. Then we have a little work to figure out the signs [3 should be positive because we want positive 1x left over].

Of course, this is a lot of tedium. The quadratic equation gives a formula to “autofactor” any quadratic equation [i.e., anything with an x^2 term].

As we get to higher powers [x^3, x^4] it becomes really hard to compute by hand, and we usually rely on computer systems to factor the equations.

This is great. I totally get that when you use the analogy of the “error”, what your referring to is that multiplication that results in a zero is a unique mathematical anomaly. You’re targeting the “Zero Product Property” as a stable point of reference in a sea of numbers.

I think what’s more difficult about factoring though, is intuitively figuring out the whole “reverse FOIL” thing. For me, it’s always been this thing where I just sit and debunk the magic combination of values that both multiply to the coefficient, and also add to the constant. Usually I descend into brute force, and while that’s fine for destroying my entire weekend, and I browse through the infinite permutations that confront me and taunt me in my homework, this literally kills me on exams with time limits.

Any shortcuts to that?

I understand the math, but this article’s use of the word “error” is confusing to me. What does it mean in this context? What’s erroneous about the equations?

I have 4 years of a math major behind me and 6 years of a science PhD, and I don’t understand what you mean by “error”.

Why not just say, you want to have zero on one side because of the special way that zero works with multiplication?

@James, yonemoto: Thanks, I need to clarify: the error is the difference between the current state and the desired state.

For example, if x = 1, we have

current state = 1^2 + 1 = 2
desired state = 6
error = 2 - 6 = -4

i.e., if we attempt to use x = 1, our error will be -4 [we will be 4 less than our target value].

If we try to use x = 4, our error is 16 + 1 - 6 = 9. We overshot our goal.

The point of factoring is to easily figure out where our error will be zero, which means we’re exactly at the desired state. Thanks for the comments, I’ll clarify!

> multiply to the coefficient, and also add to the constant

Strike that, reverse it.

@Sim (you like Wayne’s world too? :)): Great question. I think the reverse FOIL thing comes down to pattern recognition. Realistically, we usually do manual factorization on nice, premade homework problems that have nice clean solutions. In the real world we let computers factorize for us.

After doing enough problems, you start to see that factorizations work out like this:

\$\$(x + a)(x + b) = x^2 + ax + bx + ab\$\$

Notice that the a*b term has no “x’s” involved. So going backwards, if you see a regular number by itself (like -15, in x^2 + 2x - 15), you start wondering what “a” and “b” could result in -15 (remember, a and b can be negative too):

• -1 and 15
• -15 and 1
• -3 and 5
• -5 and 3

The next step is to find the combination that has a difference of 2 (since we need a 2x term left over). In this case, it’s +5 and -3. So we get

\$\$(x + 5)(x - 3) = x^2 + 5x - 3x - 15 = x^2 + 2x -15\$\$

That’s the process that goes into my head. It is brute force to some extent, and we can use the quadratic formula to blast through any equation automatically.

Honestly, most homework problems will be set up nicely so you can recognize the factors pretty quickly [what a and b have the difference we need?].

This is great. Thank you very much for the perspective. Please continue to post more insight in math!

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