Factoring

[Adron]

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.

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

[Eibro]

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.

[CrAzY]

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

I've proved the formula ;-)

[Banana fanna fo fanna]

Yes, but that doesn't work in reverse

[AntiVirus]

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