Given $f(x)=6 x^{3}+5 x^{2}-12 x+4$, answer the following.
Part: 0 / 2
Part 1 of 2
Factor $f(x)$, given that $\frac{1}{2}$ is a zero.
\[
f(x)=
\]

Step 1 ：Given that \(f(x)=6 x^{3}+5 x^{2}-12 x+4\) and \(\frac{1}{2}\) is a zero of \(f(x)\), we know that \(x - \frac{1}{2}\) is a factor of \(f(x)\).

Step 2 ：We can perform polynomial division of \(f(x)\) by \(x - \frac{1}{2}\) to find the other factor.

Step 3 ：The quotient from the polynomial division is \(6x^2 + 8x - 8\). This means that \(f(x)\) can be factored as \((x - \frac{1}{2})(6x^2 + 8x - 8)\).

Step 4 ：However, we can further factor the quadratic expression \(6x^2 + 8x - 8\) as \(2(3x^2 + 4x - 4)\), which can be further factored as \(2(3x - 2)(x + 2)\).

Step 5 ：Therefore, the complete factorization of \(f(x)\) is \((x - \frac{1}{2})2(3x - 2)(x + 2)\).

Step 6 ：Final Answer: \(\boxed{(x - \frac{1}{2})2(3x - 2)(x + 2)}\)