Factor the polynomial function $$ f(x)=4 x^{3}-19 x+15 $$ The zeros are $$ f(x)=\square $$

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To factor the polynomial function $$ f(x) = 4x^3 - 19x + 15 $$ and find its zeros, follow these steps:

### 1. **Find Possible Rational Roots**

Using the **Rational Root Theorem**, possible rational roots are factors of the constant term divided by factors of the leading coefficient. For $$ f(x) = 4x^3 - 19x + 15 $$:

- **Constant term:** 15 → ±1, ±3, ±5, ±15
- **Leading coefficient:** 4 → ±1, ±2, ±4

Possible rational roots: $$ \pm1, \pm\frac{1}{2}, \pm\frac{1}{4}, \pm3, \pm\frac{3}{2}, \pm\frac{3}{4}, \pm5, \pm\frac{5}{2}, \pm\frac{5}{4}, \pm15, \pm\frac{15}{2}, \pm\frac{15}{4} $$

### 2. **Identify an Actual Root**

Test $$ x = 1 $$:

$$
f(1) = 4(1)^3 - 19(1) + 15 = 4 - 19 + 15 = 0
$$

So, $$ x = 1 $$ is a root.

### 3. **Factor Out $$ (x - 1) $$**

Use **synthetic division** to divide $$ f(x) $$ by $$ (x - 1) $$:

$$
\begin{array}{r|rrrr}
1 & 4 & 0 & -19 & 15 \
\hline
 & 4 & 4 & -15 & 0 \
\end{array}
$$

The quotient is $$ 4x^2 + 4x - 15 $$. So,

$$
f(x) = (x - 1)(4x^2 + 4x - 15)
$$

### 4. **Factor the Quadratic $$ 4x^2 + 4x - 15 $$**

Look for two numbers that multiply to $$ 4 \times (-15) = -60 $$ and add to $$ 4 $$. The numbers are $$ 10 $$ and $$ -6 $$.

Rewrite the quadratic:

$$
4x^2 + 10x - 6x - 15 = 2x(2x + 5) - 3(2x + 5) = (2x - 3)(2x + 5)
$$

### 5. **Complete Factorization and Identify Zeros**

Thus, the fully factored form of $$ f(x) $$ is:

$$
f(x) = (x - 1)(2x - 3)(2x + 5)
$$

**Zeros of $$ f(x) $$:**

- $$ x = 1 $$
- $$ x = \frac{3}{2} $$
- $$ x = -\frac{5}{2} $$

### **Final Answer:**

$$
f(x) = (x - 1)(2x - 3)(2x + 5)
$$

## Zeros:
$$
x = 1,\ \frac{3}{2},\ -\frac{5}{2}
$$
The polynomial $$ f(x) = 4x^3 - 19x + 15 $$ factors to $$ (x - 1)(2x - 3)(2x + 5) $$. The zeros are $$ x = 1 $$, $$ x = \frac{3}{2} $$, and $$ x = -\frac{5}{2} $$.

Factor the polynomial function
The zeros are

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