To find a polynomial $Q(x)$ of degree 3 with integer coefficients, we need to account for the given zeros:

• $-7$ is one zero.
• $1 + i$ is another zero.

Since the polynomial has integer coefficients, the complex conjugate $1 - i$ must also be a zero. This ensures that the coefficients remain real and integer (as non-real roots of polynomials with real coefficients occur in conjugate pairs).

Step 1: Write the polynomial in factored form

The zeros of the polynomial are $-7$, $1 + i$, and $1 - i$. Therefore, the polynomial can be written as:

$Q(x) = (x + 7)(x - (1 + i))(x - (1 - i))$

Step 2: Simplify the complex conjugate terms

First, simplify the product of $(x - (1 + i))$ and $(x - (1 - i))$. This is a difference of squares:

$(x - (1 + i))(x - (1 - i)) = (x - 1)^2 - (i)^2 = (x - 1)^2 + 1$

So, this becomes:

$(x - 1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2$

Step 3: Multiply by the remaining factor

Now, multiply this result by the remaining factor $(x + 7)$:

$Q(x) = (x + 7)(x^2 - 2x + 2)$

Use distribution (expand the product):

$Q(x) = x(x^2 - 2x + 2) + 7(x^2 - 2x + 2)$

$Q(x) = x^3 - 2x^2 + 2x + 7x^2 - 14x + 14$

Step 4: Combine like terms

$Q(x) = x^3 + 5x^2 - 12x + 14$

Thus, the polynomial with integer coefficients that satisfies the given conditions is:

$Q(x) = x^3 + 5x^2 - 12x + 14$

Would you like more details on any part of the solution?

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Here are some related questions:

1. How do you find polynomials with more complicated roots?
2. What happens if a zero has multiplicity greater than 1?
3. How do you use synthetic division to verify polynomial roots?
4. How do complex conjugates affect the factorization of a polynomial?
5. How can polynomials with rational coefficients be factored?

Tip: Always remember that non-real roots of polynomials with real coefficients come in conjugate pairs.