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"Find all the complex roots of the following polynomials

A) $S(x)=135x^4 -324x^3 +234x^2 -68x+7$, knowing that all its real roots belong to the interval $(0.25;1.75)$

B)$M(x)=(x^3 -1+i)(5x^3 +27x^2 -28x+6)$ "

Well, in A) I don't know how to use the given information about real roots. I mean, I know that I can apply Bolzano but I don't think that's very useful. To find the complex roots I should have some information about a complex root in particular so that I could use Ruffini, but this is not the case.

And in B) I know that $(x^3 -1+i)$ is giving me some information related to a complex root, but that "^3" bothers me. If it wasn't there, I would know that $1-i$ is a root...

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  • $\begingroup$ You can find all roots of $S(x)$ by using the rational root theorem. You can find the roots of $x^3-1+i$ by using De Moivre's theorem and the roots of $5x^3+27x^2-28x+6$ by using the rational root theorem. $\endgroup$ – Varun Vejalla Jul 5 '19 at 3:52
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – saulspatz Jul 5 '19 at 3:52
  • $\begingroup$ I don't think the real roots you have provided are accurate since division does not yield a degree 2 polynomial as the Rational Root Theorem implies. x=0.333333 and x=1.4 are more accurate values, when dividing I got: $135x^2-90.000045x+15.00002$ $\endgroup$ – NoChance Jul 5 '19 at 4:11
  • $\begingroup$ @NoChance oh but they're not the roots, the real roots are between those values $\endgroup$ – AaronTBM Jul 5 '19 at 4:15
  • $\begingroup$ OK, so I guess the function is $\left(135x^2-90.000045x+15.00002\right)\left(x-0.33333\right)\left(x-1.4\right)$ You could easily find all the roots now. $\endgroup$ – NoChance Jul 5 '19 at 4:17
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Since $\frac{1}{3}$ is a root of $S$, we obtain: $$S=135x^4-324x^3+234x^2-68x+7=$$ $$=135x^4-45x^3-279x^3+93x^2+141x^2-47x-21x+7=$$ $$=(3x-1)(45x^3-93x^2+47x-7)=$$ $$=(3x-1)(45x^3-15x^2-78x^2+26x+21x-7)=$$ $$=(3x-1)^2(15x^2-26x+7)=(3x-1)^2(15x^2-5x-21x+7)=(3x-1)^3(5x-7).$$ Since $\frac{3}{5}$ is a root of $5x^3+27x^2-28x+6$, we obtain: $$5x^3+27x^2-28x+6=5x^3-3x^2+30x^2-18x-10x+6=(5x-3)(x^2+6x-2)=$$ $$=(5x-3)((x+3)^2-11)=(5x-3)(x+3-\sqrt{11})(x+3+\sqrt{11}).$$ Also, $$\sqrt[3]{1-i}=\sqrt[6]2\sqrt[3]{\cos315^{\circ}+i\sin315^{\circ}}=$$ $$=\sqrt[6]2(\cos(105^{\circ}+120^{\circ}k)+i\sin(105^{\circ}+120^{\circ}k)),$$ where $k\in\{0,1,2\}$.

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I’ll deal with (B) only. You have probably seen that $\frac35$ is a root, so that $x-\frac35$ is a factor, and when you divide $5x^3 +27x^2-28x+6$ by that, you get a $\Bbb Q$-irreducible quadratic, which you’ll have to use the Quadratic Formula on.

The factor $x^3 - (1-i)$ actually is much easier. You’re looking for the three cube roots of $1-i=\sqrt2\bigl(\cos(-45^\circ)+i\sin(-45^\circ)\bigr)$, since the point $(1,-1)$ is $\sqrt2$ units from the origin and on the ray $45^\circ$ clockwise from the $x$-axis. You need the cube root of $\sqrt2$, which you can write $2^{1/6}$ or $\sqrt[6]2$. The angles that triple to $-45^\circ$ are $-15^\circ$, $105^\circ$, and $225^\circ$. So your three linear factors of $x^3-(1-i)$ are $x-\bigl(\sqrt[6]2(\cos\theta+i\sin\theta)\bigr)$, for $\theta$ taking each of the explicit angle values I mention above.

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