(a) Find the rational zeros and then the other zeros of the polynomial function \( f(x)=x^{3}-133 x+132 \); that is, solve \( f(x)=0 \). (b) Factor \( f(x) \) into linear factors. (a) Select the correct choice below and fill in any answer box(es) within your choice. (Type integers or fractions. Use a comma to separate answers as needed. Type exact answers, using radicals as needed. Express complex numbers in terms of \( i \).) A. There are only rational zeros and the zeros are B. The rational zeros are \( \square \) and the other zeros are \( \square \). C. There are no rational zeros. The other zeros are (.

To find the rational zeros of the polynomial \( f(x) = x^3 - 133x + 132 \), we can employ the Rational Root Theorem. The possible rational zeros are factors of the constant term \(132\) over the leading coefficient \(1\). Therefore, the candidates are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 11, \pm 12, \pm 22, \pm 33, \pm 66, \pm 132\). Testing \(x=1\), we find \( f(1) = 1 - 133 + 132 = 0\), so \(x=1\) is a rational zero. Using synthetic division of \(f(x)\) by \(x-1\), we find: \[ \begin{array}{r|rrrr} 1 & 1 & 0 & -133 & 132 \\ & & 1 & 1 & -132 \\ \hline & 1 & 1 & -132 & 0 \end{array} \] This gives us \(f(x) = (x-1)(x^2 + x - 132)\). We can find the other zeros by solving \(x^2 + x - 132 = 0\) using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=1, c=-132\). This results in: \[ x = \frac{-1 \pm \sqrt{1 + 528}}{2} = \frac{-1 \pm \sqrt{529}}{2} = \frac{-1 \pm 23}{2} \] Calculating this gives us the other zeros \(x = 11\) and \(x = -12\). Thus, the rational zeros are \(1\), \(11\), \(-12\) and putting it all together: B. The rational zeros are \(1, 11, -12\) and the other zeros are none (since all zeros are rational). For factoring \(f(x)\), we've already factored it into linear factors: \[ f(x) = (x - 1)(x - 11)(x + 12) \]