To find the zeros of the polynomial function $f(x) = x^3 + 2x^2 - x - 2$, we need to solve $f(x) = 0$. Let’s proceed step by step:

Step 1: Factorization by grouping

The polynomial is: $f(x) = x^3 + 2x^2 - x - 2.$ Group the terms to factorize: $f(x) = (x^3 + 2x^2) - (x + 2).$

Factor each group: $f(x) = x^2(x + 2) - 1(x + 2).$

Factor out the common factor $(x + 2)$: $f(x) = (x + 2)(x^2 - 1).$

Step 2: Factorize further

The quadratic $x^2 - 1$ is a difference of squares: $x^2 - 1 = (x - 1)(x + 1).$

Thus, the fully factored form of the polynomial is: $f(x) = (x + 2)(x - 1)(x + 1).$

Step 3: Find the zeros

Set each factor equal to zero:

1. $x + 2 = 0 \implies x = -2$,
2. $x - 1 = 0 \implies x = 1$,
3. $x + 1 = 0 \implies x = -1$.

Final Answer:

The zeros of the polynomial are: $x = -2, \, -1, \, 1.$

Correct Options:

• $-2$
• $-1$
• $1$

Let me know if you'd like detailed verification for these solutions or further explanation.

Follow-Up Questions:

1. What is the method to solve higher-degree polynomials without factorization?
2. How do you verify whether a given value is a zero of a polynomial?
3. Can synthetic division help in finding polynomial zeros? How?
4. What are some examples of irreducible polynomials over real numbers?
5. How do the multiplicities of zeros affect the graph of a polynomial?

Tip: When dealing with polynomials, always try grouping terms or synthetic division if direct factorization isn’t obvious.