Hi

 Given the zero: x=2, of the polynomial: x^3-4x^2+21x-34=0, find the others.



 x^3-4x^2+21x-34 = (x-2)(x^2 - 2x + 17)

 x^2 - 2x + 17 = 0    |Using standard form: ax^2+bx +c = 0

 



  

    x = 1 + 4i

and x = 1 - 4i 

Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
 We are given x = 2 is one of the zeros, so we know that x - 2 is one of the factors of the polynomial 

We can divide the polynomial by x - 2 to get a quadratic:

x^3-4x^2+21x-34/x-2 

Performing the division gives 

Now use the quadratic formula to get the other two zeros:



The other two roots are complex, and are equal to:

x = 1 + 4i, and x = 1 - 4i 

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