x^3 + x^2 + ax + b = 0
Given -2+i and -2-i are roots, show that 3 is the real root.

Question

x^3 + x^2 + ax + b = 0
Given -2+i and -2-i are roots, show that 3 is the real root.

anonymous · Answer

Expansion of Vieta gives you:
$$\Large r_1 +r_2 +r_3 = -1 $$
So in your case:

$$\Large (-2+i)+(-2+-i)+r_3=-1 $$
or $$\Large r_3=3 $$

anonymous · Answer

Thank you but I don't think that is how I'm wanted to solve it because the final question is Find the sum of the three roots. That does come out to be -1 but I don't think they'd ask you to find the third root by doing it like that and then ask you to find the sum later on anyway.

anonymous · Answer

That formula also isn't on the syllabus so I don't think it would come up in an exam.

anonymous · Answer

In this case, synthetic division.

anonymous · Answer

You're gonna need to expand on that :P

anonymous · Answer

Care to explain how you do that?

anonymous · Answer

Well, the only way I can wrap this problem around my head in a different approach is to assume that $a, b $ are real, which seems obvious by the problem given. Then the question should be, given roots $r_1, r_2$ figure out what $a, b$ are and then verify that $x=3$ is a root.

anonymous · Answer

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