Maybe if we all put our minds together we can have a mathematical breakthrough.

Here's my dilemma. Back in high school, I had just switched schools so I couldn't be in any of the available classes for a whole semester. I got my hands on an intermediate algebra book and started reading through it. I figured this would be a good precursor to my precalculas class coming up.

I had passed the NYS Regent Math B exam with decent scores, but there were still so many details left out of my training. I came across polynomials in my reading and thought "This Looks Easy." The first thing I came across was:

\frac{-b \pm \sqrt(b^2-4ac)}{2a}

I had seen it before, as I am sure all of you have. Then I found the proof for it in the next few days, I will try to reproduce it here.

ax^2 + bx + c = 0

subtract c

ax^2 + bx = -c

divide by a

x^2 + \frac{b}{a}x = -\frac{c}{a}

complete the squre

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}

Simplify

(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}

Square Root of both

x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}

X by itself:

x = \frac{-b \pm \sqrt(b^2-4ac)}{2a}

Now given all of this, why can't we do the same thing for 3rd order polynomials.

ax^3 + bx^2 + cx + d = 0

ax^3 + bx^2 + cx = -d

x^3 + \frac{b}{a}x^2 + \frac{c}{a}x = -\frac{d}{a}

At this point I get lost. I would think you would take an x out of the left, however, that doesn't do much. You would think that there would be something you can add to both sides without an x in it so that it can be factored. I've done a bit of reading on this and they say there are proofs proving that it is impossible to solve this. The only thing I can think of is goofing around with imaginary numbers and what not. I can't think of anything to add to both sides of the equation to complete the "square."

x(x^2 + \frac{b}{a}x + \frac{c}{a}) = -\frac{d}{a}

To solve a cubic function, I use the factor theorem.

But I don't know a specific formula that gives the solutions of x. :(

From what I have read, there is none, and it is impossible for it to exist. However, I have never read any sensible proofs that prove that it is impossible. I am proposing a possible way to solve it algebraically.

Hmmm

ax^3 + bx^2 + cx + d = 0

ax^3 + bx^2 + cx = -d

Hmm.. factorise ax?

ax(x^2 + \frac{b}{a}x + \frac{c}{a}) = -d

Hmm... complete the square from within?

ax\((x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a}\) = -d

Hmm, expand?

ax(x+\frac{b}{2a})^2 - \frac{b^2}{4a}x + cx = -d

I'm getting nowhere... isn't there an 'complete the cubic' XD

Let's get on the reverse order then.

(x+ a)^3 = x^3 + 3ax^2 + 3a^2x + a^3

Getting backwards... by dividing the term in x^2 by 3, the third one by 3a and the cube root of the independent term... given that all the three operations give a constant 'a', that can be done... at least, that's for a perfect cubic.

Now, how do we find the remainder in case of a non-perfect cubic? :(

You are asking if there is a formula for finding the roots of a cubic as there is with the quadratic formula?

Yes, there is. It is called Cardano's formula and it is messy.

See here:

The Cubic Formula (http://www.math.vanderbilt.edu/~schectex/courses/cubic/)


Galois, Abel, or some other big brain from the past, proved there is no formula for a quintic. So forget about one for a polynomial of fifth degree. As for a quartic formula (fourth degree), it is quite daunting. Google quartic formula if you're interested.

Phew! That's really long!

I don't think I'd come up with this soon :eek:

Funny thing is, I have spent a lot of time on vanderbilt. Do you know if there is a proof for this formula? I would love to see how it was put together.

Scour around the internet and you can find it. Google 'cubic formula'. This was derived back in the 16th century, I believe. There's all kind of stuff out there.
I assume others have access to Google.

See here: http://www.sosmath.com/algebra/factor/fac11/fac11.html

You have to check out this site for the quartic formula. Talk about a mess... WHEW!!

http://myyn.org/m/article/quartic-formula/

Ew... my head is spinning :p

Thanks galactus! :) It was good to know that there is indeed a formula to find the roots of polynimials up to the 4th power.

If you're interested in the fun stuff, here is a site with Abel's proof that a quintic does not have solutions in radicals. That is, a quintic does not have a 'quintic formula' whereby to find the roots as we do with quadratics, cubics, and quartics.

Fermat's Last Theorem: Abel's Impossibility Proof (http://fermatslasttheorem.blogspot.com/2008/10/abels-impossibility-proof.html)

*cough*
I'll just bookmark this page for later :o