Question: Write the factored form of the polynomial function with real coefficients, a lead coefficient of 1, and zeros of $-4$, $-3$, $3$, and $1$.
1. **State the problem:**
We need to write the factored form of a polynomial function $y(x)$ with real coefficients, leading coefficient $1$, and zeros at $-4$, $-3$, $3$, and $1$.
2. **Recall the formula:**
If a polynomial has zeros at $r_1, r_2, \ldots, r_n$ and leading coefficient $a$, then the polynomial can be written as:
$$y(x) = a(x - r_1)(x - r_2) \cdots (x - r_n)$$
Since the leading coefficient is $1$, we have:
$$y(x) = (x - (-4))(x - (-3))(x - 3)(x - 1)$$
3. **Substitute the zeros:**
$$y(x) = (x + 4)(x + 3)(x - 3)(x - 1)$$
4. **Simplify by pairing conjugates:**
Notice $(x + 3)(x - 3)$ is a difference of squares:
$$ (x + 3)(x - 3) = x^2 - 9 $$
5. **Rewrite the polynomial:**
$$ y(x) = (x + 4)(x - 1)(x^2 - 9) $$
6. **Multiply $(x + 4)(x - 1)$:**
$$ (x + 4)(x - 1) = x^2 - x + 4x - 4 = x^2 + 3x - 4 $$
7. **Final factored form:**
$$ y(x) = (x^2 + 3x - 4)(x^2 - 9) $$
This is the factored form with real coefficients and leading coefficient $1$.
**Answer:**
$$ y(x) = (x + 4)(x + 3)(x - 3)(x - 1) $$