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The roots of $x^4+4x^3+5x^2+2x+2=0$, one root being $-1+i$ are what?

please solve this problem, i need the process of solution

qaphla
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user250822
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  • See https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem – lab bhattacharjee Jun 26 '15 at 17:49
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    It would be very helpful to know what your thoughts on this problem. People will be more responsive if you give some ideas and not just ask them to solve problems for you. – Michael Burr Jun 26 '15 at 17:51
  • Hint: As $-1+i$ is a root and the coefficients of your polynomial are real, we must have that $-1-i$ is also a root. Hence the polynomial can be divided by the quadratic $(x+1 + i)(x+i-i) = (x+1)^2 + 1$. That will reduce your problem to solving a new quadratic. – Simon S Jun 26 '15 at 17:51
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    $-1 + i$ is not a root of your polynomial... – achille hui Jun 26 '15 at 17:53
  • I think the polynomial should be$x^4+4x^3+5x^2+2x-2=0$ – Karl Jun 26 '15 at 18:23

2 Answers2

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When $x = (-1 + i)$, $x^4+4x^3+5x^2+2x+2 = 4$. Since $x^4 = -4$, I will assume that the correct polynomial in question is $$2x^4+4x^3+5x^2+2x+2.$$ Since one root is $-1 + i$, it follows that its conjugate, $-1 - i$, is also a root, and so $(x+1-i)(x+1+i)$ is a factor of the polynomial. Note that $(x+1-i)(x+1+i)$ is the difference of $(x+1)^2$ and $i^2$; $(x+1-i)(x+1+i) = x^2+2x+2$ . Using long division we can see that $$\frac {2x^4+4x^3+5x^2+2x+2}{x^2+2x+2} = 2x^2 + 1$$ The solutions of $x^2 = -\frac 12$ are $\sqrt{-1/2} = \pm \frac {i}{\sqrt{2}} = \pm \frac {\sqrt{2}i}{2}$. Thus the roots of $2x^4+4x^3+5x^2+2x+2 = 0$ are $$x = -1+i, -1-i, \frac {\sqrt{2}i}{2}, -\frac {\sqrt{2}i}{2}$$

986
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Hint: If $-1+i$ is a root, then $-1-i$ is also a root. Now do polynomial long division by $(x+1-i)(x+1+i)$, then solve the remaining quadratic.

However, it seems like $-1+i$ is actually not a root of your polynomial. Did you make an error typing it up?

Teoc
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