16 (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 $$ f(x) $$ into linear factors. (a) Select the correct choice below and fill in any answer box(es) within your choice. (Type an exact 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.) A. There are only rational zeros and they are B. There is only one rational zero, $$ \square $$, and the other zeros are C. There are no rational zeros. The other zeros are

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Let $$ f(x)=2x^3+3x^2+18x+27 $$.

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**Step 1. Find the Rational Zero**

Using the Rational Root Theorem, any rational zero is of the form  
$$
\pm \frac{p}{q},
$$  
where $$ p $$ divides the constant term $$ 27 $$ and $$ q $$ divides the leading coefficient $$ 2 $$. Thus, the candidates are  
$$
\pm 1,\, \pm 3,\, \pm 9,\, \pm 27,\, \pm \frac{1}{2},\, \pm \frac{3}{2},\, \pm \frac{9}{2},\, \pm \frac{27}{2}.
$$

Testing $$ x=-\frac{3}{2} $$:

$$
f\!\left(-\frac{3}{2}\right)
= 2\!\left(-\frac{3}{2}\right)^3 + 3\!\left(-\frac{3}{2}\right)^2 + 18\!\left(-\frac{3}{2}\right) + 27.
$$

Compute step by step:
- $$\left(-\frac{3}{2}\right)^3 = -\frac{27}{8}$$ and then $$ 2\!\left(-\frac{27}{8}\right) = -\frac{27}{4}$$.
- $$\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$$ and then $$ 3\!\left(\frac{9}{4}\right) = \frac{27}{4}$$.
- $$18\!\left(-\frac{3}{2}\right) = -27$$.

Thus,
$$
f\!\left(-\frac{3}{2}\right)
= -\frac{27}{4} + \frac{27}{4} -27 + 27 = 0.
$$

So, $$ x=-\frac{3}{2} $$ is a rational zero.

---

**Step 2. Factor Out the Rational Zero**

Since $$ x=-\frac{3}{2} $$ is a zero, we can factor $$ x+\frac{3}{2} $$ from $$ f(x) $$. Performing synthetic division with $$-\frac{3}{2}$$ on the coefficients $$ 2,\, 3,\, 18,\, 27 $$:

1. Bring down the $$ 2 $$.

2. Multiply $$ 2 $$ by $$-\frac{3}{2}$$ to get $$-3$$; add to $$ 3 $$ to obtain $$0$$.

3. Multiply $$ 0 $$ by $$-\frac{3}{2}$$ to get $$0$$; add to $$ 18 $$ to obtain $$18$$.

4. Multiply $$18$$ by $$-\frac{3}{2}$$ to get $$-27$$; add to $$ 27 $$ to obtain $$0$$.

Thus, the quotient polynomial is  
$$
2x^2+18.
$$

We can factor out the common factor $$2$$ from the quotient:
$$
2x^2+18 = 2\left(x^2+9\right).
$$

So,  
$$
f(x)=\left(x+\frac{3}{2}\right) \cdot 2\left(x^2+9\right).
$$

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**Step 3. Find the Other Zeros**

Set the remaining quadratic factor $$ x^2+9 = 0 $$:
$$
x^2 = -9.
$$
Taking square roots,
$$
x = \pm 3i.
$$

Thus, the zeros of $$ f(x)=0 $$ are:
- The rational zero: $$ x=-\frac{3}{2} $$,
- The complex zeros: $$ x=3i $$ and $$ x=-3i $$.

---

**Step 4. Factor $$ f(x) $$ into Linear Factors**

The factor $$ x+\frac{3}{2} $$ can be rewritten by multiplying by 2 as:
$$
x+\frac{3}{2}=\frac{1}{2}(2x+3).
$$

Thus, rewriting the entire factorization and incorporating the common factor 2, we have:
$$
f(x)=\left(x+\frac{3}{2}\right) \cdot 2\left(x^2+9\right)
=\frac{1}{2}(2x+3) \cdot 2\left(x^2+9\right)
=(2x+3)(x^2+9).
$$

Now, factor $$x^2+9$$ over the complex numbers as:
$$
x^2+9 = \left(x-3i\right)\left(x+3i\right).
$$

Therefore, the factorization into linear factors is:
$$
f(x)=(2x+3)(x-3i)(x+3i).
$$

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**Final Answers:**

(a) Option B: There is only one rational zero, $$-\frac{3}{2}$$, and the other zeros are $$3i$$ and $$-3i$$.

(b) The factorization into linear factors is:  
$$
(2x+3)(x-3i)(x+3i).
$$
(a) There is only one rational zero, $$-\frac{3}{2}$$, and the other zeros are $$3i$$ and $$-3i$$. (b) The factorization into linear factors is: $$ (2x+3)(x-3i)(x+3i). $$

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