1)) Which polynomial has zeros at $$ x=-4, x=-2, x=1 $$, and $$ x=3 $$ ? $$ (x+4)^{2}(x-1)^{3} $$ $$ (x-4)^{2}(x+1)^{3} $$ $$ (x-4)(x-2)(x+1)(x+3) $$ $$ (x+4)(x+2)(x-1)(x-3) $$

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To determine which polynomial has zeros at $$ x = -4, x = -2, x = 1 $$, and $$ x = 3 $$, let's analyze each option:

1. **Option A:** $$ (x + 4)^{2}(x - 1)^{3} $$
   - **Zeros:**
     - $$ x = -4 $$ (with multiplicity 2)
     - $$ x = 1 $$ (with multiplicity 3)
   - **Missing Zeros:** $$ x = -2 $$ and $$ x = 3 $$

2. **Option B:** $$ (x - 4)^{2}(x + 1)^{3} $$
   - **Zeros:**
     - $$ x = 4 $$ (with multiplicity 2)
     - $$ x = -1 $$ (with multiplicity 3)
   - **Missing Zeros:** $$ x = -4, x = -2, x = 1, $$ and $$ x = 3 $$

3. **Option C:** $$ (x - 4)(x - 2)(x + 1)(x + 3) $$
   - **Zeros:**
     - $$ x = 4 $$
     - $$ x = 2 $$
     - $$ x = -1 $$
     - $$ x = -3 $$
   - **Missing Zeros:** $$ x = -4, x = -2, x = 1, $$ and $$ x = 3 $$

4. **Option D:** $$ (x + 4)(x + 2)(x - 1)(x - 3) $$
   - **Zeros:**
     - $$ x = -4 $$
     - $$ x = -2 $$
     - $$ x = 1 $$
     - $$ x = 3 $$
   - **All Required Zeros Present**

**Conclusion:**
The correct polynomial that has zeros at $$ x = -4, x = -2, x = 1 $$, and $$ x = 3 $$ is:

**$$ (x + 4)(x + 2)(x - 1)(x - 3) $$**

**Answer:**  
$$ (x+4)(x+2)(x-1)(x-3) $$
$$ (x+4)(x+2)(x-1)(x-3) $$

1)) Which polynomial has zeros at , and ?

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