# Question about Factoring Equation

1. May 12, 2009

### optics.tech

Hi,

I use the $$\frac{-b \ \pm \ \sqrt{b^2 - 4ac}}{2a}$$ formula to factoring the quadratic equation. Can anyone tell me how to factoring the cubic equation? Usually I use the division method to factoring higher order than quadratic equation such as cubic, etc., by trial, one-by-one.

Cheers http://img395.imageshack.us/img395/7776/cheers.gif [Broken]

Last edited by a moderator: May 4, 2017
2. May 12, 2009

### HallsofIvy

Staff Emeritus
Well, use the cubic formula, of course! Just as you can use the quadratic formula to solve any quadratic equation, so you can use the cubic formula to solve any cubic equation and then the factors are D(x- a)(x- b)(x- c) where a, b, c are to roots of the equation.

Look at http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method

Needless to say, it is considerably harder than the quadratic formula. There is also a formula for solving fourth order equations but it is even harder- in part it involves reducing to a cubic equation and using Cardano's formula. There is no formula for polynomial equations of degree higher than 4. It was proved in the nineteenth century that there exist such equations whose solutions cannot be written in terms of radicals.

3. May 12, 2009

### epenguin

The quoted pages may look a bit formidable to you.
The way I remember it and could always construct a solution if needed is: just as you solve a quadratic by "completing the square" - i.e. by expressing it as the difference of two squares, then the solution of that difference = 0 is one square equals the other... you know the rest, so you try to express the cubic as difference of two cubes. You find that in order to do that you are led to have to solve a quadratic equation.

You find that if the cubic equation has 3 real roots you will need to use the nonreal complex roots of that quadratic to get them. They could find no way not to use these, so that led to the discovery that square roots of negative numbers are not just a silly answer to a silly question but something serious, useful and it turned out, inevitable. I.e. it was hoped to find some way to solve using just real numbers but it was proved this would be impossible. I don't know how elementary and accessible that last bit is and am curious.

Anyway the point is that what looks like a specialised problem, solving the cubic, turns out to lead to one of the most important things in mathematics, also surely one of the key steps in the process of abstraction in maths. The ideas and calculation for solving the cubic and quartic are really less difficult and more natural than the formidable-looking formulae would lead you to think.

Last edited: May 13, 2009
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