Suppose that a polynomial function of degree 4 with rational coefficients has i and (-3 + square root of 3)as zeros find the other zeros
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To find the other zeros, we can start by using the fact that if a polynomial has rational coefficients, then the complex conjugates of its complex zeros are also zeros.
Given that the zeros of the polynomial are i and (-3 + sqrt(3)), we know that their conjugates are -i and (-3 - sqrt(3)), respectively. Therefore, these complex conjugates are also zeros of the polynomial.
To find the other zeros, we can write the polynomial function using these known zeros:
(x - i)(x + i)(x - (-3 + sqrt(3)))(x - (-3 - sqrt(3)))
Now we can multiply these factors to obtain the polynomial form:
=(x^2 + 1)(x + 3 - sqrt(3))(x + 3 + sqrt(3))
To find the remaining zeros, we can set the polynomial equation equal to zero:
(x^2 + 1)(x + 3 - sqrt(3))(x + 3 + sqrt(3)) = 0
Now we solve for x by setting each factor equal to zero:
1) x^2 + 1 = 0
This equation has no real solutions, but it has two complex solutions: x = i and x = -i.
2) x + 3 - sqrt(3) = 0
Rearranging this equation, we have: x = sqrt(3) - 3.
3) x + 3 + sqrt(3) = 0
Rearranging this equation, we have: x = -sqrt(3) - 3.
Therefore, the remaining zeros of the polynomial are i, -i, sqrt(3) - 3, and -sqrt(3) - 3.