To form a polynomial with the given zeros and degree, we use the fact that the polynomial can be expressed as a product of factors corresponding to its zeros. Each zero $c$ with multiplicity $m$ contributes a factor $(x - c)^m$ to the polynomial.

Given:

• Zeros: $-9$ (multiplicity 1) and $-1$ (multiplicity 2)
• Degree: 3
• Leading coefficient: 1

Steps:

1. Write the factors for each zero:

   • For $-9$, $(x - (-9)) = (x + 9)$
   • For $-1$ (multiplicity 2), $(x - (-1))^2 = (x + 1)^2$

2. Combine the factors: $f(x) = (x + 9)(x + 1)^2$

3. Expand $(x + 1)^2$: $(x + 1)^2 = (x + 1)(x + 1) = x^2 + 2x + 1$

4. Multiply $(x + 9)$ by $x^2 + 2x + 1$: $f(x) = (x + 9)(x^2 + 2x + 1)$

   Use distributive property: $f(x) = x(x^2 + 2x + 1) + 9(x^2 + 2x + 1)$

   Expand: $f(x) = x^3 + 2x^2 + x + 9x^2 + 18x + 9$

   Combine like terms: $f(x) = x^3 + 11x^2 + 19x + 9$

Final Answer:

$f(x) = x^3 + 11x^2 + 19x + 9$

Would you like further clarification or a verification of the steps? 😊

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Related Questions:

1. How do you verify whether a given polynomial has specified zeros?
2. What happens to the polynomial if the leading coefficient changes?
3. How do multiplicities affect the graph of a polynomial?
4. Can you find a polynomial with different leading coefficients for the same zeros?
5. How do you factorize $x^3 + 11x^2 + 19x + 9$ back into its original components?

Tip:

When expanding or factoring polynomials, always double-check your arithmetic to avoid small mistakes.