To solve this problem, we need to construct a polynomial with the following properties:

1. It has rational coefficients.
2. The leading coefficient is 1.
3. It has the given zeros $0$, $2 - \sqrt{2}$, $2 + 3i$.

Step 1: Rational Coefficients

Since the polynomial must have rational coefficients:

• If $2 - \sqrt{2}$ is a zero, its conjugate $2 + \sqrt{2}$ must also be a zero.
• If $2 + 3i$ is a zero, its conjugate $2 - 3i$ must also be a zero.

Thus, the zeros of the polynomial are: $0, 2 - \sqrt{2}, 2 + \sqrt{2}, 2 + 3i, 2 - 3i.$

Step 2: Form the Polynomial

The polynomial can be constructed by multiplying the factors corresponding to these roots: $f(x) = x \cdot \big(x - (2 - \sqrt{2})\big) \cdot \big(x - (2 + \sqrt{2})\big) \cdot \big(x - (2 + 3i)\big) \cdot \big(x - (2 - 3i)\big).$

Step 3: Simplify the Factors

Real Roots:

The factors for $2 - \sqrt{2}$ and $2 + \sqrt{2}$ combine into: $\big(x - (2 - \sqrt{2})\big)\big(x - (2 + \sqrt{2})\big) = \big(x - 2 + \sqrt{2}\big)\big(x - 2 - \sqrt{2}\big).$ Using the difference of squares: $= (x - 2)^2 - (\sqrt{2})^2 = (x - 2)^2 - 2.$

Complex Roots:

The factors for $2 + 3i$ and $2 - 3i$ combine into: $\big(x - (2 + 3i)\big)\big(x - (2 - 3i)\big) = \big(x - 2 - 3i\big)\big(x - 2 + 3i\big).$ Using the difference of squares: $= (x - 2)^2 - (3i)^2 = (x - 2)^2 - (-9) = (x - 2)^2 + 9.$

Combine All Factors:

Now include $x$: $f(x) = x \cdot \big((x - 2)^2 - 2\big) \cdot \big((x - 2)^2 + 9\big).$

Step 4: Expand the Polynomial

Let $y = (x - 2)^2$. Then: $f(x) = x \cdot (y - 2) \cdot (y + 9).$ First expand $(y - 2)(y + 9)$: $(y - 2)(y + 9) = y^2 + 9y - 2y - 18 = y^2 + 7y - 18.$ Substitute back $y = (x - 2)^2$: $f(x) = x \cdot \big((x - 2)^4 + 7(x - 2)^2 - 18\big).$

Expand $(x - 2)^2 = x^2 - 4x + 4$, then substitute and simplify further if needed.

Would you like me to continue with full expansion or clarify anything? Let me know!