Find a polynomial $$ f(x) $$ of degree 3 with real coefficients and the following zeros. $$ -3,-1+i $$ $$ f(x)=\square $$

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We are given that the polynomial $$ f(x) $$ has degree 3 and real coefficients with zeros $$-3$$ and $$-1+i$$. Since the coefficients are real, the complex conjugate of $$-1+i$$, which is $$-1-i$$, must also be a zero. Therefore, the zeros are:

$$
-3,\; -1+i,\; -1-i
$$

A polynomial with these zeros can be written (up to a constant multiple) as:

$$
f(x) = a(x+3)(x+1-i)(x+1+i)
$$

We can choose $$ a = 1 $$ to obtain a monic polynomial. Now, observe that the product of the factors involving the complex numbers is:

$$
(x+1-i)(x+1+i)
$$

This expression is in the form of a difference of squares. Recall that:

$$
(a-b)(a+b) = a^2 - b^2
$$

Here, $$ a = x+1 $$ and $$ b = i $$, so the product becomes:

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

Since $$ i^2 = -1 $$, we have:

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

Now, the polynomial becomes:

$$
f(x) = (x+3)\left[(x+1)^2+1\right]
$$

Next, expand $$(x+1)^2$$:

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

Therefore:

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

Now, multiply the factors:

$$
f(x) = (x+3)(x^2+2x+2)
$$

Expanding this product:

$$
\begin{aligned}
f(x) &= x(x^2+2x+2) + 3(x^2+2x+2) \
     &= x^3 + 2x^2 + 2x + 3x^2 + 6x + 6 \
     &= x^3 + (2x^2+3x^2) + (2x+6x) + 6 \
     &= x^3 + 5x^2 + 8x + 6
\end{aligned}
$$

Thus, the polynomial is:

$$
f(x) = x^3 + 5x^2 + 8x + 6
$$
$$ f(x) = x^3 + 5x^2 + 8x + 6 $$

Find a polynomial of degree 3 with real coefficients and the following zeros.

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