$1-5.5$ ) - Question 9 of 12 inis quiz: 12 poinl(9) This question: 1 point(s) possible Submit quiz $1 \leftarrow$ Find the complex zeros of the following polynomial function. Write $f$ in factored form \[ f(x)=x^{3}-11 x^{2}+43 x-65 \] The complex zeros of $f$ are (Simplify your answer. Type ahexact answer, using radicals and $i$ as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) Use the complex zeros to factor $f$. \[ f(x)= \] (Factor completely. Type an exact answer, using radicals and $i$ as needed. Use integers or fractions for any numbers in the expression.)

Step 1 ：Given the polynomial function \(f(x) = x^3 - 11x^2 + 43x - 65\)

Step 2 ：The polynomial does not factor nicely and does not have any rational roots, so we use the cubic formula to find the roots

Step 3 ：The cubic formula is given by \(x = [q + \sqrt{q^2 + 4p^3/27}]^{1/3} + [q - \sqrt{q^2 + 4p^3/27}]^{1/3} - b/3a\)

Step 4 ：Where \(a, b, c,\) and \(d\) are the coefficients of the cubic polynomial \(ax^3 + bx^2 + cx + d = 0\), and \(p\) and \(q\) are given by \(p = c/a - b^2/3a^2\) and \(q = 2b^3/27a^3 - bc/3a^2 + d/a\)

Step 5 ：For the given polynomial, \(a = 1, b = -11, c = 43,\) and \(d = -65\)

Step 6 ：Calculating \(p\) and \(q\), we get \(p = 43 - (-11)^2/3 = 43 - 121/3 = 43 - 40.33 = 2.67\) and \(q = 2*(-11)^3/27 - (-11)*43/3 - 65 = -242.67\)

Step 7 ：Substituting \(p\) and \(q\) into the cubic formula, we get \(x = [242.67 + \sqrt{(242.67)^2 + 4*(2.67)^3/27}]^{1/3} + [242.67 - \sqrt{(242.67)^2 + 4*(2.67)^3/27}]^{1/3} + 11/3\)

Step 8 ：Solving this, we get \(x = 5, 3 + 4i, 3 - 4i\)

Step 9 ：\(\boxed{\text{Therefore, the complex zeros of the polynomial are 5, 3 + 4i, and 3 - 4i}}\)

Step 10 ：Using these zeros, we can factor the polynomial as \(f(x) = (x - 5)(x - (3 + 4i))(x - (3 - 4i))\)