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swissgirl
Group Title
How to find the complex roots of
\(x^4+x^3+3x^2+2x+2 \)
 2 years ago
 2 years ago
swissgirl Group Title
How to find the complex roots of \(x^4+x^3+3x^2+2x+2 \)
 2 years ago
 2 years ago

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dumbcow Group TitleBest ResponseYou've already chosen the best response.2
you have to factor out any real roots easiest to graph the function to obtain any real roots
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
There are no real roots though
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
its above the y axis
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let (x^2+bx+1)(x^2+cx+2) = that expression and find the value of b and c
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
ok then yeah what @experimentX said if two of the complex roots is in the form: \[x = a \pm bi\] then \[xa = \pm bi\] \[(xa)^{2} = b^{2}\] \[x^{2}+ (2a)x +(a^{2}+b^{2}) = 0\] anyway , my point is that a pair of complex roots come from a quadratic equation :
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
I imagine you could factor this by grouping and then use the quadratic formula again.
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
i tried doing the method u taught me yesterday but it wldnt go
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
\[\Large x^4+x^3+3x^2+2x+2 \\ \Large =x^4+2x^2+x^3+x^2+2(x+1)\\ \Large =x^4+2x^2+x^2(x+1)+2(x+1)\\ \Large =x^2(x^2+2)+(x^2+2)(x+1)\\ \Large =(x^2+2)(x^2+x+1)\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
I think the trick is seeing to split up the \(3x^2\) as \(2x^2+x^2\).
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
omg y is it always sooooo simple and I can never see it lol
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Thanksssss guyssss for helping me :))))
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
You're welcome.
 2 years ago
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