To factor the polynomial \( x^3 + 2x^2 - 9x - 18\), we can use the method of grouping.
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First, we can try to find a rational root using the Rational Root Theorem. Testing potential roots, we find that \( x = -3 \) is a root because: \[ (-3)^3 + 2(-3)^2 - 9(-3) - 18 = -27 + 18 + 27 - 18 = 0. \]
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Now, we can factor \( x + 3 \) out of the cubic polynomial. We can perform polynomial long division of \( x^3 + 2x^2 - 9x - 18 \) by \( x + 3 \).
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When we divide, we find that: \[ \frac{x^3 + 2x^2 - 9x - 18}{x + 3} = x^2 - x - 6. \]
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Now we can factor \( x^2 - x - 6 \): \[ x^2 - x - 6 = (x - 3)(x + 2). \]
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Therefore, the original polynomial can be factored as: \[ x^3 + 2x^2 - 9x - 18 = (x + 3)(x - 3)(x + 2). \]
So the completed factorization in the form \((x^2 - \text{(something)})(x + \text{(something)})\) is: \[ (x^2 - x - 6)(x + 3) \quad \text{or} \quad (x^2 - 3)(x + 3). \] Thus, filling in the blanks will give: \[ (x^2 - 3)(x + 2). \]