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 May 22nd, 2015, 04:47 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Roots of the complex equations Find all the roots for the following equation. $2x^4-x^3-x^2+3x+1=0$ My attempt, I factorised it to $(x+1)(2x^3-3x^2+2x+1)=0$ So I know one of its roots is -1. How to proceed then? Is there way to find the complex solution without using computer algebra system?
 May 22nd, 2015, 05:36 AM #2 Senior Member     Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 Cardano's method.
 May 22nd, 2015, 05:38 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I don't know how you're expected to find the roots. If you can find them numerically, use bisection between -1 and 0 and then divide out the approximate root; with what's left you can use the quadratic formula to get the complex roots. If you need a symbolic solution you need Cardano's formula.
 May 22nd, 2015, 06:31 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,967 Thanks: 2216 You need to solve $2x^3 − 3x^2 + 2x + 1 = 0$. Substituting $x = u + 1/2$ and dividing by 2 gives $u^3 + (1/4)u + 3/4 = 0$. Now let $3ab = -1/4$ and $a^3 + b^3 = 3/4$ and (carefully) solve these equations by obtaining and solving a quadratic equation whose roots are $a^3\!$ and $b^3\!$. It remains to solve $u^3 - 3abu + a^3 + b^3 = 0$, i.e., $(u + a + b)(u + \omega a + \omega^2b)(u + \omega^2 a + \omega b) = 0$, where $\omega$ is a complex cube root of 1, such as $-\frac12 + \frac{\sqrt3}{2}i$, and hence find $x$.

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