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Given that x=2 is a root of p(x)=x^4-3x^3+6x^2-12x+8, find the other roots (real and nonreal).

waffles  Dec 4, 2017
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x^4  - 3x^3  + 6x^2  - 12x  + 8

 

Rewrite as  :

 

x^4  - 3x^3  + 2x^2  +  4x^2 - 12x + 8

 

x^2 (x^2- 3x + 2) +  4 (x^2 - 3x + 2)

 

(x^2 + 4)  ( x^2 - 3x  + 2)

 

(x^2  + 4)  (x -2) ( x - 1)

 

The roots are    x  = 1, x = 2   and  x = ±2i

 

 

cool cool cool

CPhill  Dec 4, 2017

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