First, let's find the real zeros of the polynomial function [tex]\( f(x) = 7x^4 + 6x^3 - 78x^2 - 66x + 11 \)[/tex].
The real zeros of the polynomial [tex]\( f \)[/tex] are the values of [tex]\( x \)[/tex] that make the equation [tex]\( f(x) = 0 \)[/tex] true. Based on the calculations and simplifications:
The real zeros of [tex]\( f \)[/tex] are:
[tex]\[
x = -\sqrt{11}, -1, \sqrt{11}, \frac{1}{7}
\][/tex]
So the correct choice is:
A. The real zero(s) of [tex]\( f \)[/tex] is/are [tex]\(-\sqrt{11}, -1, \sqrt{11}, \frac{1}{7}\)[/tex].
Next, we use these real zeros to factor the polynomial [tex]\( f \)[/tex]. The zeros tell us the factors of [tex]\( f(x) \)[/tex] in the form:
[tex]\[
(x + \sqrt{11})(x - \sqrt{11})(x + 1)\left(x - \frac{1}{7}\right)
\][/tex]
We can combine [tex]\( (x + \sqrt{11})(x - \sqrt{11}) \)[/tex] into [tex]\( (x^2 - 11) \)[/tex], and [tex]\( \left(x - \frac{1}{7}\right) \)[/tex] can be rewritten as [tex]\( (7x - 1) \)[/tex] to match the polynomials' degrees and simplify:
The fully factored form of [tex]\( f(x) \)[/tex] is:
[tex]\[
f(x) = (x + 1)(7x - 1)(x^2 - 11)
\][/tex]
