Using SubstitutionDate: 10/05/98 at 21:53:52 From: Joshua Gronsbell Subject: Explanation of your Explanation of Cubic Polynomials In your section on solving cubic polynomials, you make a leap of logic that I just don't understand. Here's the explanation: "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." This much I'm okay with. Where things get complicated is the next step: "The next thing we do is to get rid of the x^2 term by replacing x with (x - e/3)." How exactly does this work? I don't understand how replacing x with (x-e/3) helps anything, or, for that matter, how it gets rid of the x^2 term. Maybe it's just me, but if you could please explain this further, it would be most appreciated. Thanks. Date: 10/06/98 at 19:49:46 From: Doctor Peterson Subject: Re: Explanation of your Explanation of Cubic Polynomials Hi, Joshua. You are referring to the Dr. Math FAQ on cubic equations, which can be found here: http://mathforum.org/dr.math/faq/faq.cubic.equations.html That explanation obviously assumed you are familiar with some of the methods you need to use to solve cubics. Substitution is a useful technique that we don't tend to teach in algebra. We often see it in calculus instead. The basic idea is that we can change to a new variable for which the equation is simpler, because some terms drop out. It's easier to follow if we don't reuse x, but call the new variable y. So let's let: x = y - e/3 and put that in place of x in our equation: x^3 + ex^2 + fx + g = 0 (y - e/3)^3 + e(y - e/3)^2 + f(y - e/3) + g = 0 y^3 - e*y^2 + e^2*y/3 - e^3/27 + e*y^2 - 2e^2*y/3 + e^3/9 + f*y - ef/3 + g = 0 y^3 +(f - e^2/3)y +(g - ef/3 + 2e^3/27) = 0 Do you see how the y^2 term dropped out? You might like to think about why that happened, and why y - e/3 was chosen as the substitution in order to make that happen. Pretty smart, huh? (Of course, e, f, and g would be numbers derived from your specific problem, so this would look simpler than it does here.) - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/