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May 22nd, 2015, 05:47 AM   #1
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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?
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May 22nd, 2015, 06:36 AM   #2
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Cardano's method.
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May 22nd, 2015, 06:38 AM   #3
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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.
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May 22nd, 2015, 07:31 AM   #4
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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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