Write a polynomial function of least degree with integral coefficients that have the given zeros of -1, 1, i√6

Using the Factor Theorem, the polynomial function of least degree is:
[tex]f(x) = x^4 + 5x^2 - 6[/tex]
The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:
[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]
In which a is the leading coefficient.
In this problem, the real zeros are: [tex]x_1 = -1, x_2 = 1[/tex].
There is a complex zero at [tex]x_3 = \sqrt{6}i[/tex], hence there will also be a zero at it's conjugate, that is, [tex]x_4 = -\sqrt{6}i[/tex].
Then, considering a leading coefficient of a = 1, the polynomial is:
[tex]f(x) = (x + 1)(x - 1)(x - \sqrt{6}i)(x + \sqrt{6}i)[/tex]
Considering that i² = -1:
[tex]f(x) = (x^2 - 1)(x^2 + 6)[/tex]
[tex]f(x) = x^4 + 5x^2 - 6[/tex]
You can learn more about the Factor Theorem at https://brainly.com/question/24380382