Answer:
remaining zeros: {-1/2, √3}
Explanation:
Given the polynomial function f(x) = 2x³ +x² -6x -3 has one zero at x = -√3, you want the remaining zeros.
Conjugate zeros
A polynomial with rational real coefficients and zeros that are roots or complex numbers will have all such roots in conjugate pairs.
The root that is the conjugate of -√3 is +√3.
Product of roots
The product of the roots of an odd-degree polynomial will be the opposite of the ratio of the constant to the leading coefficient. Here, that means the product of roots is ...
-(-3/2) = 3/2
We already know that two of the three roots are ±√3, so their product is -3. The remaining root will be ...
(3/2)/(-3) = -1/2
The remaining zeros of f(x) are √3 and -1/2.
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Additional comment
When the degree of the polynomial is even, the constant term will be the product of the roots. This odd/negative, even/positive behavior comes from the fact that each zero (q) corresponds to a factor (x -q).
The product of these binomial factor constants is the constant in the standard-form polynomial. It will have a negative sign if there are an odd number of factors.