factorising

Is there a trick to factoring the quadratic:

ax(to the power of four) + bx³ + cx² + dx + e

My actual problem is: z^4 + 2z^3 - 4z^2 - 2z + 3.

All help appreaciated! :)

The rational root theorem is the only way I know.  The root of the equation

z^4+2z^3-4z^2-2z+3=0 are either 3, -3, 1, or -1.  

Trial and error gives a root of z=1.  So that means the expression is divisible by z-1.

(z^4+2z^3-4z^2-2z+3)/(z-1)=z^3+3z^2-z-3.  That factors into (z+1)(z^2-1)

So the final answer is (z+1)^2(z-1)^2

I don't think there's a way to factor a ... whatever its called.  Why do we call ax^2+bx+c quadratic if it only has 3 terms?  I have heard there is some way to get roots of a cubic, but I don't know how.

thank you! :)

Solving quartics can be done in general, for any values of the coefficients, but it is a way  more complicated than the formula for the quadratic that everyone learns at school (well, probably most fail to learn it).

This was a problem which inspired a lot of effort from 16th century mathematics to solve it (and the cubic). For a while their methods were considered so valuable they were kept secret.

If you want to learn about this (rather advanced) topic, it would be best to first learn about the general solution of the cubic, as this is required as part of the solution of the quartic. Historically, I think a way of converting a quartic problem to a cubic problem was discovered, and then a little later a general solution for the cubic was discovered.

Are there any closed form solutions yet of pentic, hextic, septic, octic, novtic, dectic, etc.?  n-tic?

No. There are only closed solution for quadratic, cubic and quartic equations. Polynomial equations on higher degree can be solved only in special cases, not generally. (And actually, solution of cubic, for example, can also be difficult, if there is the third root of complex number. So the solution in closed form does not solve always the whole problem.)

Actually, tseta is not quite correct. While there is no closed solution of the quintic in radicals (i.e. some formula involving rational powers of combinations of the coefficients), there is a closed solution using the definition of a  Bring radical as a solution of  X^5 + X + c = 0. But the formulas are so complicated they were not written down until 2004, as far as I know, and take 3 pages for a single formula! There is another way of expressing the solutions using a more exotic sort of number called icosians relating to the symmetries of the icosohedron. This was discovered by Felix Klein, well known for his discovery of the Klein bottle.

The reason for the sudden change in the nature of the solutions between quartics and quintics is a key topic in the history of algebra. The set of all permutation of n objects (the objects are the roots) is an algebraic object called a "group" with n! elements. The crucial thing is that for n = 1 to 4 this group can be "decomposed" into cyclic groups of orders 2 and 3. These are associated with square roots and cube roots (eg, a 4th root is simply a square root of a square root, and the cyclic group of order 4 breaks into two cyclic groups of order 2).

But when you get to the permutations of 5 objects, this group can only be broken down into a cyclic group of order 2 and an irreducible object with 60 elements (which happens to be the rotations of the icosohedron).

All permutation groups of order more than 4 can only be broken down in a similar way, so one would think there might be solutions of the sextic, heptic polynomials relating to the groups A(6), A(7) etc for higher order polynomials. There are indeed general solutions of the quintic and all higher polynomials using a class of multi-variable functions called generalised hypergeometric functions, taking variables based on the coefficients of the polynomial. But these solutions are inherently different to the single variable radical functions (i.e. n'th roots) that are enough up to the quartic, and the single variable Bring radicals that suffice for the quintic.

It's interesting how a problem like factoring polynomials has been around so long, while these days hardly anyone is interested in this problem anymore. I ran into the same thing when I was writing a proof and needed to know which numbers can be written as a sum of two, three or four integral squares. Mathematicians like Euler even studied such problems, while these days it is not even taught at university anymore.

I remember from a class on the history of mathematics that indeed, at some point in history, people found a way to reduce a quartic polynomial to a cubic polynomial, and then also found a way to solve cubic polynomials. However, this reduction and these "general solution" formulas are very ugly.

Since the most common way to run into a "factor a degree d>2 polynomial"-problem is that a teacher asks you to do this in class, and since most teachers make sure at least one of the roots is easy to find, you can just use trial and error in most cases. Or you can write out the product (x+r1)(x+r2)(...)(x+rk) and try solving the (nonlinear) equations you get for the coefficients of the polynomials. For example (x+a)(x+b)(x+c) = x^3 + (a+b+c) x^2 + (ab+ac+bc) x + abc, so for the polynomial x^3 + 14 x^2 + 53 x + 40 you'd get the equations a+b+c = 14, ab+ac+bc = 53, abc = 40 with solutions 1,5,8.

In answer to pawn_slayer666's interesting question as to why a quadratic has that name (one might expect it to have a name relating to 2, the highest power), the reason seems to be that "quadra" is a square in Latin, so the name is consistent with the name "cubic" for 3rd order equations. It just happens that the square shape is named after its number of sides in Latin, (4 is "quattura").

For those with a big appetite for these sorts of problems, you should read up on some Galois theory, which (loosely speaking) deals with what types of polynomials can be solved using radicals. I should warn you though it gets a bit hairy if you don't have your abstract algebra fresh in memory..

It is true that to show the connections between Galois theory and solving polynomials rigorously is rather technical (I think I recall the need for a "butterfly lemma" from when I studied this in the early 80s?). However,  to get the central idea, all you need is to understand basic finite group theory with the fundamental facts about normal subgroups, and perhaps an empirical look at A5 to show it is a simple group. (For those who have not studied this stuff, a "simple group" is one with no normal subgroups, i.e. it cannot be decomposed. Simple groups are to group theory a bit like prime numbers are to number theory)