factorising x^4 .... (1 Viewer)

[kurt.physics]

Is there any equation for factorising (similar to the difference of 2 cubes) for algebraic sentences with x^4?

 

[Iruka]

Well, you can treat x^4 - y^4 like the difference of two squares and go from there.

For x^4 + y^4 there is a tricky factorisation:

x^4 + y^4 = x^4 + 2x^2y^2 + y^4 - 2x^2y^2
= (x^2+y^2)^2- 2x^2y^2
= (x^2 + sqrt(2)xy + y^2)(x^2 - sqrt(2)xy+y^2)

 

[tommykins]

Not that I can think of, unless you use a dummy variable such as m = x^2.

 

[Slidey]

Is there any equation for factorising (similar to the difference of 2 cubes) for algebraic sentences with x^4?

Yep, there certainly is.

Here it is: http://planetmath.org/encyclopedia/QuarticEquation.html

I dare you not to scream when you look at it.

I suggest you never bother learning it, and examine its existence from a purely abstract context. E.g. for asking why a formula exists for solving degree 1, 2, 3, 4 polynomials, but not degree 5 or more.

As others have said, some tricks exist for some types:

First, check evident roots, mainly 0, 1 and -1 (through the p(k)=0 means (x-k) is a root test). Otherwise:

1) x^4-x^3+x-1=0 (common factors)
x^3(x-1)+(x-1)=0
(x^3+1)(x-1)=0 (difference of two cubes)
(x+1)(x-1)(x^2-x+1)=0 (use quadratic formula)

2) x^4+4x^3+6x^2+4x-15=0 (complete the quartic)
(x^4+4x^3+6x^2+4x+1)=16
(x+1)^4=16
x+1=2, -2 (plus two others, but find them after division by x-1 and x+3, through quadratic formula)
It still works for e=3 for example, too, but you get a nasty fourth root.

Another example of completing the quartic:
x^4+4x^3+4x+1=0
x^4+4x^3+6x^2+4x+1=-6x^2
(x+1)^4+6x^2=0
((x+1)^2+sqrt(6)x)((x+1)^2-sqrt(6)x)=0 (then use quadratic formula)

This method has limited appeal due to the fact that it's hard to notice a perfect quartic, let alone find one with ONLY an altered x^2 coefficient OR constant (but not both or anythinse else).

3) x^4+2x^2+1=0
y=x^2
y^2+2y+1=0
y=-1
x=i, -i.

4) x^4+2x^3+x^2-1=0
x^2(x^2+2x+1)-1=0 (difference of two squares)
(x(x+1)+1)(x(x+1)-1)=0 (use quadratic formula)

5) x^4+x^3+x^2-x+1=0 (divide by x^2)
x^2+x+1-1/x+1/x^2=0
z=x-1/x (square this to find z^2)
(x^2+2+1/x^2)+(x-1/x)-1=0
z^2+z-1=0 (quadratic formula to solve for z, then quadratic formula again to solve for x)

It actually works whenever you have:
ax^4+bx^3+cx^2+bmx+am^2=0
E.g.: the above case was:
a=1, b=1, c=1, m=-1, but any of these can be any integer and it still works.

 

[buchanan]

The usual way to solve an asymmetric quartic is to form a resolvent cubic first, solve that, and then 2 quadratics.

where the cubic in y can be solved using the <a href="http://users.tpg.com.au/nanahcub/cubic.gif">cubic formula</a>.

This is a much easier method than the formula on the planetmath website.

 

[Slidey]

Haha true that, and some alternative solutions are listed on Wikipedia:

http://en.wikipedia.org/wiki/Quartic_equation#Alternative_methods

However, it's worth noting that except in semi-trivial cases, none of the methods available (including your own buchanan) are especially fast or easy (both to use and remember), and all are plagued by the possibility of numerical errors. These formulae exist, in my opinion, primarily for computer algorithms.

Quasi-symmetric equations
It should be noted that for the quasi-symmetric equations:
ax^4+bx^3+cx^2+bmx+am^2=0
The substitution is not z=x-1/x, but z=x+m/x. I forgot to mention this in my earlier post.

This is an important case to remember because quasi-symmetric cases turn up a lot, at least in at a HSC level. E.g.:
x^4+5x^2+1=0
5x^4+3=0
3x^4+2x^3+12x+16=0
x^4+x^3+x^2-x+1=0