Factoring cubics?

[SilverSprite]

Hey anybody here wanna give me a couple of ways to factor cubics, and i'm not talking about difference of cubes or sum of cubes. Just how to factor cubics.

[SilverSprite]

Here too!

[A$$Bandit]

In aid of what exactly? On the off chance (and it seems to be quite a small one at that) that you're genuinely asking for some help (and not just posting another "who can/can't answer this question" thread like you do most of the time):

f(x) = ax³ + bx² + cx + d

Say f(x) = x³ - x² - x - 2 and you wanna factorise it.

First you might want to plug a few small integers into f(x) and see what you get. Say you guess 1, -1, 2, and see that f(2) = 0. This tells you that (x - 2) is a factor of f(x). You could guess more numbers, but I'm telling you here that you won't find any more that work because I've chosen f(x) for this reason. All cubics have at least one real root, some have 3. But because some only have 1, you can't go guessing for more real factors because there might not be any more, and you might be wasting your time.

So you found out that (x - 2) is a factor. Next step...

f(x) = (x - 2)(... + ... + ...) <please ignore the ...., it's just there to separate the brackets.

You have to fill in the second bracket so that when multiplied, they make f(x). With cubics, this is dead easy because it's always a quadratic and thus you only have 3 values to fit in correctly.

Well:
The original expression (f(x)) had a 1x³ in, so you know that the first term of the bracket must be x². It's the only way to get a 1x³ out of multiplying them. So you fill it in:

f(x) = (x - 2)(x² + ... + ...)

Next you look at the constant on the end of f(x), because much like the x³ there's only one way to get this when the brackets are expanded. The constant is -2, so you gotta add something onto the end of that bracket which when multiplied by -2, makes -2. Clearly the answer is 1 - fill it in.

f(x) = (x - 2)(x² + ... + 1)

Both of those terms are easy because of the singular ways to get them in expansion. However the term in x is a bit more tricky because you have to consider that it can be made two ways: x*ConstantAtEndOfSecondBracket AND -2*xTerm. Look at f(x) again. You need to make -x overall. You can see from the incomplete brackets that you are already gonna get +1x on expansion, so you need to balance that by taking off 2x, i.e. the product of -2 (from the first bracket) and the coefficient of the term you're choosing has to be -2 (so you size the overall term in x down to equal -x). The answer here would be +x, because that means overall you get x*constant(1) + x*-2, which equals -x. So fill that in:

f(x) = (x - 2)(x² + x + 1)

Now you see if you can factorize that quadratic into two more brackets. By inspection, you see that you can't, as there are no two factors of +1 which add to make +1. If it was factorizable you'd just shove the two brackets in place of that quadratic bracket and leave it at that, fully factorized. But because it doesn't just factorize by inspection, you'd use the quadratic formula to see if there are any real roots, and you'd see that b² - 4ac is negative, indicating there are no more real roots (just as well you stopped guessing then!), and that the quadratic can't be factorized. So that's it, f(x) is factorized fully as stated in the most previous identity.

Well Mr. Sprite...
If you needed help then I hope that sorted it out for you. If you were just asking another pointless question then it seems you've got your come-uppance.

[Guv]

Except for some special cases, you cannot easily factor a cubic without solving for one of its roots. Once you know a root, it is simple.

There is an analytical algorithm for finding roots. Do a search for Cardano or Tartaglia. The basic idea is that you divide by the leading coefficient to get the following form.

x³ + bx² + cx + d

Then substitute x = y - b/3 in the above, getting the following form,

y³ + py + q

Then substitute y = z - p/3z and you get a quadratic in z³, which can be solved using the formula for quadratic roots. Once you know z, you can work backwards to get y & x.

Except for people taking certain archaic mathematics courses, nobody solves cubics using the above algorithm. Various successive approximation methods are used. Newton-Raphson is a very good method. It is easy to program and easy to just do on a hand calculator.

If you search this forum for Newton, you will find that method described in at least one thread.

[SilverSprite]

Yes there is a cubic formula. I am familiar with it and thanx to Bugz i can prove it . But it is really really long. It involves depressing the cubic ax^3+bx^2+cx+d=0 and i do not recommend it. If you guys would like furthur info then go to www.sosmath.com and search for cubic formula. The proof is there.

[DavidHooper]

Hi Sprite,
I've got a Excel worksheet which can factorise all cubics with integer roots. Not much help, I know, but I'll post it up if it'll help you.

[sql_lall]

One word of advice: When you get this way to factorize cubics, don't come up with the cubic equivalent of the quadratic equation. It is SO long, and BIG (i.e. takes up room to write) that i suggest u just do it in the order they say.

http://mathworld.wolfram.com/CubicEquation.html

(Sorry, i know it's long winded, but it's from the best maths site: www.mathworld.wolfram.com Go there for almost anything)

[Kalkewl8ter]

Guessing at the factors by inspection will give you an answer much more quickly than finding roots, or other methods.

(_x+_)*(_(x^2)+_x+_)

Fill in the blanks.

Or, use synthetic division.

[Guv]

Kalkewl8ter: Lots of luck filling in the blanks if the real root is not an integer.

[A$$Bandit]

Well regardless of any formula, the quickest way I've found of solving/factorizing cubics or higher equations (when you know the roots/brackets are all integers) is the way I've said above.