View Full Version : finding zeros of a 3rd degree polynomial

PaulKraemer

05-30-2011, 10:24 AM

Hi,

Can anyone help me find the zeros of the following equation:

f(x) = 2x^3 + x^2 - 20x + 1

I used the rational zero therom to determine that the *possible* zeros are the following:

+/- 1

+/- 1/2

I then tried using synthetic division to test if any of the above *possible* zeros actually are zeros. For all of the possibilities (+1, -1, +1/2, -1/2), I get a remainder in my synthetic division.

If anyone could help me with this, I'd really appreciate it.

Thanks in advance,

Paul

galactus

05-30-2011, 11:04 AM

The roots to this are not 'nice'.

There is a formula for cubic just as there is to quadratics. Just enter in your a,b,c,d.

In this case, a=2, b=1, c=-20, d=1

The formula is rather ugly by comparison. But, it's plug and chug.

It's called Cardano's formula

See here:

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Denis

05-30-2011, 02:18 PM

As galactus says, the roots are uglier than the 7 witches of MacBeth combined!

Go see: http://www.numberempire.com/equationsolver.php

IF you're learning cubics, I wonder why you're given this!

Subhotosh Khan

05-30-2011, 08:00 PM

As galactus says, the roots are uglier than the 7 witches of MacBeth combined!

Now that is ugly .... super ugly....

PaulKraemer

05-31-2011, 01:46 PM

Thanks galactus, Denis, and Subhotosh,

I actually got this exercise in a calculus chapter on finding local extrema where they ask you to sketch the graph of the function. In the example problems in this chapter, they calculate the local extrema and puts those points on the graph. Next, they find the zeros and put those points on the graph. Then they calculate f(0) and put that point on the graph. With all of these points, it is easy to sketch the graph pretty accurately. In all of the example problems, it was easy to find the zeros.

Obviously, it is not a trivial task to find the zeros of the function I posted this question about. I am guessing that they don't expect you to through such an effort with Cardano's theorem just to have a few extra points to plot on the graph. I am happy now that at least I understand how to do it if I had to.

Thanks again,

Paul

galactus

05-31-2011, 02:26 PM

You can use the Intermediate Value theorem. This is guess and check until you hone in on the roots.

2x^{3}+x^{2}-20x+1=0

Try a guess of x=2. This results in -19

Try x=3, this gives 4

It changes from negative to positive, so the root is between 2 and 3.

Keep honing in until you get it as accurate as you want. If plotting, a would say a decimal place or two should suffice.

Try x=-4, this gives -31

Try x=-3, this gives 16

So, there is a root between -3 and -4

There is one more closer to 0.

PaulKraemer

05-31-2011, 04:00 PM

Thanks galactus,

That is a big help!

Kind regards,

Paul

Hi, Paul.

To add to what has already been suggested, MathWorld also has some good information regarding the Cubic formula:

Cubic Formula on MathWorld

http://mathworld.wolfram.com/CubicFormula.html

And so does Wikipedia:

Cubic Equation Information on Wikipedia

http://en.wikipedia.org/wiki/Cubic_equation

Both of these pages present exact formulae for calculating exact, algebraic, results.

On the other hand, if you just want some quick numbers, there are also several online free solvers for numerically solving a cubic polynomial, for example,

Cubic Equation Solver

http://www.akiti.ca/Quad3Deg.html

Hope this helps,

Tom

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