To find the zeros of the polynomial function \( f(x) = x^3 + 2x^2 - 9x - 18 \), we can use the Rational Root Theorem and synthetic division or other methods (like factoring or the quadratic formula if applicable) to test for possible rational roots.
Step 1: Find Rational Roots
The Rational Root Theorem suggests that the possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, the constant term is \(-18\) and the leading coefficient is \(1\). The factors of \(-18\) are \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\).
We can test these possible roots using synthetic division or direct substitution into the polynomial.
Step 2: Testing Possible Roots
- Testing \( x = -3 \): \[ f(-3) = (-3)^3 + 2(-3)^2 - 9(-3) - 18 = -27 + 18 + 27 - 18 = 0 \] Hence, \( x = -3 \) is a root.
Step 3: Synthetic Division
Now that we have found one root, we will perform synthetic division of \( f(x) \) by \( x + 3 \).
-3 | 1 2 -9 -18
| -3 3 18
-----------------
1 -1 -6 0
The quotient is \( x^2 - x - 6 \).
Step 4: Factor the Quadratic
Now we can factor \( x^2 - x - 6 \): \[ x^2 - x - 6 = (x - 3)(x + 2) \]
Step 5: Find All Zeros
So the factored form of \( f(x) \) is: \[ f(x) = (x + 3)(x - 3)(x + 2) \]
Setting each factor equal to zero gives the zeros:
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x - 3 = 0 \) → \( x = 3 \)
- \( x + 2 = 0 \) → \( x = -2 \)
Zeros and Multiplicity
The zeros are:
- \( x = -3 \) with multiplicity 1
- \( x = 3 \) with multiplicity 1
- \( x = -2 \) with multiplicity 1
Step 6: Behavior at the Zeros
Since all the zeros have odd multiplicity (1), at each zero, the graph:
- Crosses the x-axis at \( x = -3 \)
- Crosses the x-axis at \( x = 3 \)
- Crosses the x-axis at \( x = -2 \)
Final Answer
The zeros are: \[ \text{The zero(s) is/are } -3, 3, -2 \]
Behavior
- For \( x = -3 \): Crosses the x-axis
- For \( x = 3 \): Crosses the x-axis
- For \( x = -2 \): Crosses the x-axis