SOLUTION: How to find polynomial with solutions: (0, 2-√3, 3) ? So far I did (x-0)(x+3i)(x-3i), but am not sure what to do for 2-√3. Thanks!

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Question 1065196: How to find polynomial with solutions: (0, 2-√3, 3) ?
So far I did (x-0)(x+3i)(x-3i), but am not sure what to do for 2-√3. Thanks!
Found 2 solutions by ikleyn, rothauserc:
Answer by ikleyn(52796)   (Show Source): You can put this solution on YOUR website!
.
The correct question is: 

    How to find a polynomial with integer coefficients, having the roots 0, 2- sqrt(3) and 3 ?

Such a polynomial together with the root 2-sqrt(3), must have the root 2+sqrt(3), too.
So, the polynomial must have factors 
     x                 for the root 0;

     (x-(2-sqrt(3)))   for the root  (2-sqrt(3));

     (x-(2+sqrt(3)))   for the root  (2+sqrt(3));

     (x-3)             for the root 3.

Thus the polynomial must be a multiple to

     p(x) = x*(x(x-(2-sqrt(3)))((x-(2+sqrt(3)))*(x-3)

You may write it in more convenient form.



Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
You have the product of
:
x(x-3)(x-2+square root(3))
:
Note that -2+square root(3) = -0.2679 is approximately -0.27
:
Therefore,
:
x(x-3)(x-0.27)
:
x(x^2 -3.27x +81)
:
x^3 -3.27x^2 +81x
:
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