Cubic PolynomialDate: 6/27/96 at 14:23:46 From: Anonymous Subject: Cubic Polynomial Sketch the graphs of y=ax^3, y=ax^3+mx, y=ax^3-mx for some a,m>0. Note that all of these graphs have a point of symmetry at the origin. Show that every cubic polynomial function y=ax^3+bx^2+cx+d can be expressed in the form y=a(x-h)^3+m(x-h)+k for some h,k; thus the graph of every cubic polynomial has a point of symmetry. Date: 9/12/96 at 11:26:24 From: Doctor Ken Subject: Re: Cubic Polynomial Hello! This is our standard answer for the question "What are the general solutions to cubic and quartic polynomial equations?" This actually gives a description of how to solve the cubic, but we don't describe how to do the quartic. It's just too darn messy. Contained in the method for solving the cubic equation is a transformation (a variable substitution, which is essentially the step you ask for too - substitute x-h for x) that will eliminate the squared term. Okay, here's how you do the cubic. Let's say you have the equation ax^3 + bx^2 + cx + d = 0. The first thing you do is to get rid of the "a" out in front by dividing the whole equation by it. Then we get something in the form of x^3 + ex^2 + fx + g = 0. The next thing we do is to get rid of the x^2 term by replacing x with (x - e/3). That will give us something of the form of x^3 + px + q = 0. This is much easier to solve, although it's still kind of hard and it took a lot of really smart people a really long time to do. Let's introduce two new variables, t and u, defined by the equations u - t = q and tu = (p/3)^3. If you're really weird, you might want to know what t and u actually are, in terms of p and q. Well, here they are: -3q +- Sqrt{4p^3 + 27 q^2} 27q +- Sqrt{108p^3 + 729q^2) t = ------------------------- , u = ---------------------------- 6 Sqrt{3} 54 Where that +- is either + or - for BOTH of the equations at the same time. You can't have one as a + and the other as a -. Anyway, x = CubeRoot{t} - CubeRoot{u} will be a solution of x^3 + px + q = 0. Verify this result now (DON'T use the explicit formulas for t and u, but just plug CubeRoot{t} - CubeRoot{u} into the cubic polynomial), and make sure you see why it works. To find the other two solutions of our cubic polynomial (if there are any real ones) we could divide x^3 + px + q by its known factor (x - CubeRoot{t} + CubeRoot{u}), getting a quadratic equation that we could solve by the quadratic formula. As you can see the complexity is much greater than it is in the Quadratic Formula. To try to go backwards and come up with a closed form for the Cubic Formula in terms of the original a, b, c, d would be a real pain. If you'd like a nice reference about this problem, as well as some stuff about the Quartic, see if you can lay your hands on the 3-volume set called "Mathematical Thought from Ancient to Modern Times" by Morris Kline. It gives the math and the history together, which is pretty cool. -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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