Find the real zeros of $\mathrm{f}$. Use the real zeros to factor $\mathrm{f}$.
\[
f(x)=x^{3}-17 x^{2}+81 x-81
\]
The real zero(s) of $\mathrm{f}$ is/are $9,4 \pm \sqrt{7}$
(Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
Factor $\mathrm{f}$.
\[
f(x)=
\]
(Factor completely. Type ah exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression.)

Step 1 ：The real zeros of a polynomial are the x-values for which the polynomial equals zero. These are also the x-intercepts of the graph of the polynomial.

Step 2 ：The real zeros of the function \(f(x)=x^{3}-17 x^{2}+81 x-81\) are \(9, 4 - \sqrt{7}, 4 + \sqrt{7}\).

Step 3 ：Once we find the zeros, we can factor the polynomial by setting each factor equal to zero and solving for x.

Step 4 ：The factored form of the function is \((x - 9)(x - (4 - \sqrt{7}))(x - (4 + \sqrt{7}))\).

Step 5 ：\(\boxed{The real zeros of the function f(x)=x^{3}-17 x^{2}+81 x-81 are 9, 4 - \sqrt{7}, 4 + \sqrt{7}. The factored form of the function is (x - 9)(x - (4 - \sqrt{7}))(x - (4 + \sqrt{7})).}\)