Solving Cubics (3rd Degree Polynomials)
Date: 12/15/96 at 11:28:34 From: Kenny Walden Subject: Solving Cubics (3rd Degree poly's) Is there a straightforward way to solve problems of the type ax^3+bx^2+cx+d=0 ? I know it's possible to do it guess and check, but at competitions you don't have that much time. A formula would be most helpful, but a step-by-step process of factoring would also be very useful.
Date: 12/16/96 at 09:38:16 From: Doctor Jerry Subject: Re: Solving Cubics (3rd Degree poly's) Hi Kenny, The formula for a cubic is much more complex than the formula for solving a quadratic. There is no known "step-by-step process of factoring," other than trying the possible rational roots (for polynomials with integer coefficients) by synthetic division. That is, if you want to solve the cubic: 2x^3+x^2+x-1=0 then (from a theorem), IF this equation has a rational root, it must be in the list (1,-1,1/2,-1/2). You can test these by synthetic division. Writing down 2 1 1 -1 and testing 1/2 gives the numbers 2 2 2 0. The 0 means that 1/2 is a zero and that: 2x^3+x^2+x-1 = (x-1/2)(2x^2+2x+1). If it happens that the polynomial equation has no rational roots, then a numerical method or Cardan's Method must be used - for a complete discussion, look at this answer in our archives: http://mathforum.org/dr.math/problems/handel.6.20.96.html I hope this answers your question. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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