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Cubic Equations
By Anonymous on July 22, 1998:

I am reading the book Mind Tools by Rudy Rucker (Penguin) and he says on page 128 that any cubic can be turned into the form k(term in x3 + Ux). Why is there no term in x2?

The same he says is true with higher polynomials: if the order is a quartic then no term is required in x3.

I'd be grateful for help on these problems as I have a very lively class of IB students doing Higher Maths and no one in my department can answer !

Thank you

By Curran Roller on July 28, 1998:

The reason for changing a "general" cubic equation (Ax3 + B x2 + Cx + D = 0) into a "depressed" cubic equation (kx3 + kUx = N) goes back to the 15th century, when Italian mathematicians were first searching for a way to solve all real cubics, however general. I recommend the book "Journey Through Genius" by William Dunham for a fascinating coverage of the both the mathematics and the interpersonal intrigue which accompanied this chapter in the history of mathematical discovery.

The upshot of it all, and the reason for creating depressed cubics and quartics such as the example you gave, is that depressed cubics and quartics are very solvable, and so the ability to replace general polynomials with the depressed ones is the key to finding their roots. It turns out that there is such a thing as the "cubic formula", similar to the general quadratic formula, which gives a solution to any cubic equation (although, compared to the quadratic formula, it's way more complicated).

Quartics may also be solved using a general algorithm; in both cases sometimes the solution you get may look very complicated (radicals within radicals, and such). I hope you can find a copy of the Dunham book, it's tremendous literature.

By Eva on July 31, 1998:

Dear David,
It is easy to see how a cubic polynomial can be transformed into another cubic with no x2 term. Suppose the cubic is ax3 + bx2 + cx + d, and call this p(x). Then for any number w,
p(x-w)
= a(x-w)3 + b(x-w)2 + c(x-w) + d
= ax3 + [b - 3aw]x2 + ....
= ax3 + Cx + D
if we take w = b/3a. Note that a is not zero (else p is not a cubic).

More generally, if p(x) = axn + bxn-1 + .... then
p(x-w)
= axn + [b-naw]xn-1 + ....
= axn + terms in xn-2 and lower order
if we take w = b/na.

You state that any cubic can be transformed into the form ax3 + bx; to do this you have to consider p(x-u)+v. Choose u as above, and then choose v so that this new polynomial has no constant term.

By Gareth McCaughan on September 16, 1998:
"It is easy to see how a cubic polynomial can be transformed into another cubic with no x2 term."

"You state that any cubic can be transformed into the form ax3 + bx;"

Just a remark: the fact that you can do this transformation proves a simple but nice theorem:
The graph of any cubic has rotational symmetry of order 2. (Proof: the transformations above correspond to translating the graph horizontally and vertically. They result in a cubic, namely y=ax3 + bx, whose graph obviously has rotational symmetry of order 2 about the origin (because rotation by 180 degrees is the same as negating both x and y coordinates, and that obviously takes points on the graph to points on the graph. QED.)

Of course this isn't in the least important; but it's pretty...