# Calculating complex roots

"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...

• 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. – Varun Vejalla Jul 5 '19 at 3:52
• wolframalpha.com/input/… – saulspatz Jul 5 '19 at 3:52
• 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: $135x^2-90.000045x+15.00002$ – NoChance Jul 5 '19 at 4:11
• @NoChance oh but they're not the roots, the real roots are between those values – AaronTBM Jul 5 '19 at 4:15
• 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. – NoChance Jul 5 '19 at 4:17

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 $$k\in\{0,1,2\}$$.
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.