# Thread: Polynomials with powers of three

1. ## Polynomials with powers of three

Hello Id just like to know if there is a way to solve polynomials that use x^3

2. Simple answer. No, Polynomials does have a "solution". Equations do.

There is a general solution the general cubic equation. You don't want to know it. This is why we invent numerical methods.

Hello Id just like to know if there is a way to solve polynomials that use x^3
There are several ways to solve a cubic equation. Those that you would see in a math class usually can be solved either by factoring, or by the rational root theorem (which you can look up). Many can't be solved by these methods, but require more complicated methods or the "cubic formula". If you google "cubic equation", you can see explanations of these methods, e.g. on Wikipedia.

4. Originally Posted by Dr.Peterson
There are several ways to solve a cubic equation. Those that you would see in a math class usually can be solved either by factoring, or by the rational root theorem (which you can look up). Many can't be solved by these methods, but require more complicated methods or the "cubic formula". If you google "cubic equation", you can see explanations of these methods, e.g. on Wikipedia.
I'd like to supplement this very good answer.

First, as tkhunny pointed out, there is a cubic formula (similar to the quadratic formula) that is very lengthy and looks quite ugly. If you can avoid errors, that formula will give you an exact answer in radicals, which can be turned into a decimal approximation using a scientific calculator. Unlike tkhunny, I see no reason not to use a formula that will give an exact answer that, if you own a decent calculator, can easily be reduced to a very accurate numerical approximation.

Second, unless you run across cubic equations a lot, do not memorize the cubic formula. If you need it, you can look it up. For most math homework problems, you can use a graphing calculator if you have one. If not, you can probably use factoring or (worst case) the integer or rational root methods to turn the cubic into a product of a linear term and quadratic term like (x - p)(ax^2 + bx + c) and then use the quadratic formula to turn it into a(x - p)(x - q)(x - r). By the zero product property, p, q, and r are the zeroes of the polynomial. The advantage of these methods over the cubic formula is that it is very easy to make an error with the cubic formula.

5. Originally Posted by JeffM
Unlike tkhunny, I see no reason not to use a formula that will give an exact answer that, if you own a decent calculator, can easily be reduced to a very accurate numerical approximation.
You answered your own question. If you have a decent calculator, it will give you the "very accurate numerical approximation" without any need to see the relatively useless exact answer.

OP: Same for Quartic Equations. There is a general solution, but again, I would be stunned if you ever actually needed it. It's one of those do-it-once-so-that-you-believe things and then move on with your life and with your education.

6. Originally Posted by tkhunny
You answered your own question. If you have a decent calculator, it will give you the "very accurate numerical approximation" without any need to see the relatively useless exact answer.

OP: Same for Quartic Equations. There is a general solution, but again, I would be stunned if you ever actually needed it. It's one of those do-it-once-so-that-you-believe things and then move on with your life and with your education.
I am not being antagonistic. I am curious how you would use a calculator to get a highly accurate approximation of the roots of a cubic, perhaps Newton's method?

If you are talking about a graphing calculator, we can also stop wasting time on the quadratic formula, the rational root theorem, and factoring. They are all obsolete tools of a pre-computer age. If we deny ourselves graphing calculators, I suspect the cubic formula is more efficient computationally than Newton's method (unless perhaps you can start with a very good initial estimate), but you may be thinking about a better method of approximating than Newton's method.

7. I never take anyone as being antagonistic.

Count Operations.

Cardano: Just for the Real Roots: Square roots(3), Cube roots(3), Divisions(9), Multiplications(4). In order to get a good approximation from this historically-amazing formula, there is an awful lot of as-close-as-you-want calculation going on. None of these operations is trivial. Of course, your desired accuracy must be maintained in each of those 19 separate calculations. Precision (rounding and machine) error is not absent, here. Different installations may approach the fractional exponents in different ways.

Newton: Multiplications(9) and division(1) - per iteration, of course. In my example that I used just to count operations, it took 9 iterations to achieve 15 decimal places of accuracy. Had I tried at all to refine my starting location, perhaps through visual means, I can reduce this to 4 iterations. At some point, this numerical method is less burdensome to the CPU.

Certainly, there are or can be even faster methods. Newton's Quadratic Convergence is pretty hard to beat. Do we REALLY need more than 15 decimal places for most things?

There's always a trade-off. Sometimes, we trade for elegance. Cardano certainly does that.

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