Yeah...I have a math test tomorrow, and factoring is a bitch. What's the fastest/easiest way to do it? Can someone enlighten me here? It's just that I'm usually good at math, but factoring seems like sort of a guessing game to me.

[quote author=St0rm.iD link=board=36;threadid=5537;start=0#msg46985 date=1078187115]
Yeah...I have a math test tomorrow, and factoring is a bitch. What's the fastest/easiest way to do it? Can someone enlighten me here? It's just that I'm usually good at math, but factoring seems like sort of a guessing game to me.
[/quote]

Here are some good things to memorize:

x[sup]2[/sup] - a[sup]2[/sup] ==> (x + a) (x - a)
EX: x[sup]2[/sup] - 9 ==> (x + 3) (x - 3)

a[sup]2[/sup] + 2ab + b[sup]2[/sup] ==>(a + b)[sup]2[/sup]
EX: x[sup]2[/sup] + 6x + 9 ==> (x + 3)[sup]2[/sup]

a[sup]2[/sup] - 2ab + b[sup]2[/sup] ==>(a + b)[sup]2[/sup]
EX: x[sup]2[/sup] - 6x + 9 ==> (x - 3)[sup]2[/sup]

Edit: Accidentally posted instead of previewing. Will post more. Be patient.

You can figure out how the signs work thusly:
Ax[sup]2[/sup] + Bx + C ==> (? + ?) * (? + ?)
Ax[sup]2[/sup] + Bx - C ==> (? + ?) * (? - ?)
Ax[sup]2[/sup] - Bx + C ==> (? - ?) * (? - ?)

Other than that, the method I use is fairly easy:
(x + #1)(x + #2) = Ax[sup]2[/sup] + Bx + C
What two numbers multiplied together give you C and added together give you B? Multiply out these combinations. If it doesn't work, try the next one.

Thanks.

It seems like a guessing game at first, but after a while you begin to recognize factors almost instantly. Just practice more!

An often useful technique is finding roots (just guessing a zero or so) for the polynomial and then dividing by the corresponding term.

It is a guessing game. You can remember the very frequent ones, and easily guess the less frequent but more obvious ones.
If there's something you can't guess, just use the root formula.

Ex:
[color=yellow]x² + 10x + 6[/color]
A few seconds of thought wasted on guessing, you decide to turn to the formula.
[color=cyan]x1, x2 = (-10 ± sqrt(10² - 4 * 1 * 6)) / 2 = -5 ± sqrt(19)[/color]
Therefore:
[color=yellow]x² + 5x + 3[/color] = [color=cyan](x + 5 - sqrt(19))(x + 5 + sqrt(19))[/color]

Ex:
[color=yellow]x² + 4x + 8[/color]
A few more seconds of thought wasted on this one, and the formula seems attractive again.
[color=cyan]x1, x2 = (-4 ± sqrt(4² - 4 * 8)) = ...[/color] Oops? The value inside the square root is negative.
Therefore, [color=yellow]x² + 4x + 8[/color] cannot be factored over the reals. (It can be factored over the complex numbers, but I'm guessing you aren't asked to do this.)

Note: The two examples I gave in this post can be solved in a quicker way by a method called "completing the square". Have you learned this yet?

Edit: Forum turned the 8) into a smily.

A relevant question is: What degree of polynomials do you expect to be factoring? For second degree polynomials, the formula or completing the square are great methods. For higher degree polynomials, you'll have to use tricks, guessing, etc.

Typical solutions for polynomials of higher degrees than 2 on our tests have involved rather obvious guessable roots (0, 1, -1 etc).

Quadratic Formula, Completing the Square, and Difference of Squares are the basic ones.

Completing the square would be like x² - 9 --> (x - 3)(x + 3)
The two different signs add a 3x then subtract the 3x, effectively leaving you with "x² - 9" again.

[quote author=j0k3r link=board=36;threadid=5537;start=0#msg47061 date=1078230946]
Quadratic Formula, Completing the Square, and Difference of Squares are the basic ones.

Completing the square would be like x² - 9 --> (x - 3)(x + 3)
The two different signs add a 3x then subtract the 3x, effectively leaving you with "x² - 9" again.
[/quote]

I *Always* use the quadratic formula (unless it's an obvious one). Then you aren't guessing at all.

x=(-b +/- sqrt(b[sup]2[/sup]-4ac))/2a

This will give you two answers, like -4 and 7 for instance, then the roots are (x+4) and (x-7). If you do it like this, and it's one of the hard ones, you won't have wasted your time looking for integral factors.

[quote author=iago link=board=36;threadid=5537;start=0#msg47070 date=1078237724]
I *Always* use the quadratic formula (unless it's an obvious one). Then you aren't guessing at all.

x=(-b +/- sqrt(b[sup]2[/sup]-4ac))/2a

This will give you two answers, like -4 and 7 for instance, then the roots are (x+4) and (x-7). If you do it like this, and it's one of the hard ones, you won't have wasted your time looking for integral factors.
[/quote]

But that depends on the polynomial not having a higher degree than 2. Those never require any guessing, but if you can guess an obvious solution, sure. It's when you reach higher degrees that you need to start guessing...

That's true. There's probably Quadratic formulas for any degree if somebody bothered to invent them. I want a Quintatic or Sextatic formula :(

[quote author=iago link=board=36;threadid=5537;start=0#msg47106 date=1078256443]
That's true. There's probably Quadratic formulas for any degree if somebody bothered to invent them. I want a Quintatic or Sextatic formula :(
[/quote]NO!!! Have you never heard of Abel's Impossibility Theorem?
http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html

There are formulas for 3rd and 4th degree, both insanely complicated.

Edit: Oops, missed obvious opportunity for joke, so here it is:
I'm sure you would like a sextatic formula.
Ok, this post is now complete.

[quote author=Yoni link=board=36;threadid=5537;start=0#msg47128 date=1078264396]Have you never heard of Abel's Impossibility Theorem?[/quote]

I think it's pretty obvious that I haven't :P

We always used synthetic division to factor polynominals with a degree greater than 2. It usually took awhile to find the roots of 4th and 5th+ degree functions, but it's pretty straightforward.

Yes, we do factor over complex numbers :)

And I got an 81%...not bad...

[quote author=Eibro link=board=36;threadid=5537;start=0#msg47171 date=1078281842]
We always used synthetic division to factor polynominals with a degree greater than 2. It usually took awhile to find the roots of 4th and 5th+ degree functions, but it's pretty straightforward.
[/quote]

Well having a root it's straightforward. It's finding the root that is the only problem, that's where guesswork comes in.

There's a bit of guess work, yeah. We learned to find possible rational roots by taking factors of the constant, and dividing them by factors of the leading coefficient. Ex.
f(x) = 2x^3 + 3x^2 + 2x + 10;

c = +- 1, 2, 5, 10
l = +- 1, 2

Possible rational roots are therefore: +- 1, 1/2, 2, 5, 5/2, 10. Just keep plugging them into the eq. until you come out with 0. It's quick if you have a graphing calculator you can program the function into. Though it does get ugly when you get constants/coefficients with a lot of factors.

St0rm, you should of just down x=(-b+-root(-b+4ac))/2a

I've proved the formula ;-)

Yes, but that doesn't work in reverse :)

St0rm you in Alg I ? And If you need help with alg I just aim me at ThePsychoJoker18, math is very easy for me. But I am only in alg II, cause I am in 10th grade. So if you need help just halla ;)