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(52795)   (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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