Find the roots of the following polynomial equation..

Question

how would you solve this exercise:

Find the solutions of the following equation knowing that one of these solutions belongs to $R$:

$$x^3+(3i-2)x^2-(1+4i)x+2+i=0$$

I used the condition set in the problem and got four values of that real solution, which is a quiet weird thing. So, may be you'll show me some clever and convincing way!

Thank you in advance!!!

As you know a cubic polynomial over the complex numbers has three roots, counted according to multiplicity. If you can find a real root (a version of Rational Roots Them. applies here), then you can remove a corresponding factor to get a quadratic polynomial. Then things should be simple. — hardmath, Apr 18 '14 at 16:32

score 3 · Accepted Answer · answered Apr 18 '14 at 16:26

3

Let $a\in R$ is one of the solutions

So,we have $$a^3+(3i-2)a^2-(1+4i)a+2+i=0$$

Equating the real & the imaginary parts

$$a^3-2a^2-a+2=0\implies a^2(a-2)-(a-2)=0\iff (a-2)(a^2-1)=0$$ and $$3a^2-4a+2=0\iff $$

So, the common value of $a=?$

answered Apr 18 '14 at 16:26

lab bhattacharjee

274,582

Here was my problem, I didn't choose the common value, but was set foolishly on checking each of them. Thank you! – wonderingdev Apr 18 '14 at 16:29

score 1 · Answer 2 · answered Apr 18 '14 at 16:26

By inspection, $\;x=1\;$ is a real root, so now divide your polynomial by $\;x-1\;$ and use the quadratic formula for the roots.

That one root is real doesn't seem to be very helpful in general: what helped here is that one real root is a really easy one to check...

Find the roots of the following polynomial equation..

2 Answers2