To construct a polynomial with the given zeros, we need to account for both the given roots and any necessary conjugate roots. The given zeros are $-3$ and $-1 + \sqrt{3}$.

Step 1: Include the Conjugate Root

For the zero $-1 + \sqrt{3}$, its conjugate $-1 - \sqrt{3}$ must also be a root of the polynomial to ensure that the coefficients of the polynomial are integers.

Step 2: Write the Factors

The polynomial will have factors corresponding to each root:

• For the root $-3$, the factor is $(x + 3)$.
• For the roots $-1 + \sqrt{3}$ and $-1 - \sqrt{3}$, the factors are $(x + 1 - \sqrt{3})$ and $(x + 1 + \sqrt{3})$.

Step 3: Multiply the Conjugate Factors

First, let's multiply the conjugate factors: $(x + 1 - \sqrt{3})(x + 1 + \sqrt{3})$

This is a difference of squares: $(x + 1)^2 - (\sqrt{3})^2$ $= (x + 1)^2 - 3$ $= x^2 + 2x + 1 - 3$ $= x^2 + 2x - 2$

Step 4: Multiply All Factors

Now, multiply this result by the factor corresponding to the root $-3$: $(x + 3)(x^2 + 2x - 2)$

Leave the polynomial in factored form: $\boxed{(x + 3)(x^2 + 2x - 2)}$

This is the polynomial of least degree with integer coefficients that has the given zeros.

Would you like to see the polynomial expanded or have any further questions?

Here are 5 related questions:

1. How do you determine the conjugate of a root that involves square roots?
2. What is the difference between a polynomial in factored form and expanded form?
3. Can you explain why the product of conjugates eliminates irrational numbers?
4. What does the "least degree" of a polynomial mean?
5. How would the process change if one of the roots were imaginary?

Tip: Always remember to include conjugate pairs when forming polynomials with real coefficients to eliminate any irrational parts.