To factor the polynomial \( x^3 + 2x^2 - 9x - 18 \), we can look for potential rational roots using the Rational Root Theorem. The potential roots can be factors of the constant term (-18) divided by the factors of the leading coefficient (1).
The factors of -18 are: \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
Let's test these potential roots one by one:
- Testing \( x = -2 \): \[ (-2)^3 + 2(-2)^2 - 9(-2) - 18 = -8 + 8 + 18 - 18 = 0 \] Therefore, \( x = -2 \) is a root.
We can then perform polynomial long division or synthetic division of \( x^3 + 2x^2 - 9x - 18 \) by \( x + 2 \).
Synthetic Division:
-2 | 1 2 -9 -18
| -2 0 18
---------------------
1 0 -9 0
The resulting polynomial from the synthetic division is \( x^2 - 9 \).
So we can write: \[ x^3 + 2x^2 - 9x - 18 = (x + 2)(x^2 - 9) \]
Next, we can factor \( x^2 - 9 \) further since it is a difference of squares: \[ x^2 - 9 = (x - 3)(x + 3) \]
Putting it all together, the complete factorization of \( x^3 + 2x^2 - 9x - 18 \) is: \[ x^3 + 2x^2 - 9x - 18 = (x + 2)(x - 3)(x + 3) \]
Thus, the final factored form is: \[ \boxed{(x + 2)(x - 3)(x + 3)} \]
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