(a) Find the rational zeros and then the other zeros of the polynomial function $f(x)=2 x^{3}+3 x^{2}+18 x+27$, that is, solve $f(x)=0$.
(b) Factor $\mathrm{f}(\mathrm{x})$ into linear factors.

Step 1 ：First, we use the Rational Root Theorem to find the rational zeros of the polynomial function \(f(x)=2 x^{3}+3 x^{2}+18 x+27\). The Rational Root Theorem states that any rational root, p/q, of a polynomial with integer coefficients must have p as a factor of the constant term and q as a factor of the leading coefficient. In this case, the constant term is 27 and the leading coefficient is 2.

Step 2 ：We list out all the possible rational roots and test them by substituting them into the polynomial. If the result is zero, then it is a root of the polynomial.

Step 3 ：After finding the rational roots, we use synthetic division to reduce the polynomial and find the other roots.

Step 4 ：The rational roots of the polynomial are \(x = -\frac{3}{2}\). The other roots are \(x = -3i\) and \(x = 3i\), which are complex roots.

Step 5 ：We can use these roots to factor the polynomial into linear factors.

Step 6 ：The polynomial can be factored into linear factors as \(f(x) = (2x + 3)(x^{2} + 9)\).

Step 7 ：\(\boxed{\text{Final Answer: The rational zeros of the polynomial function } f(x)=2 x^{3}+3 x^{2}+18 x+27 \text{ are } x = -\frac{3}{2}. \text{ The other zeros are } x = -3i \text{ and } x = 3i. \text{ The polynomial can be factored into linear factors as } f(x) = (2x + 3)(x^{2} + 9)}\)