To factor the polynomial \(x^3 + 2x^2 - 9x - 18\), we can apply the factor theorem and synthetic division or polynomial long division. The first step is to find a root of the polynomial.
We can try some rational roots. Testing \(x = -3\):
\[ f(-3) = (-3)^3 + 2(-3)^2 - 9(-3) - 18 \] \[ = -27 + 2(9) + 27 - 18 \] \[ = -27 + 18 + 27 - 18 = 0 \]
So, \(x = -3\) is a root. Now we can perform synthetic division of \(x^3 + 2x^2 - 9x - 18\) by \(x + 3\):
\[ \begin{array}{r|rrrr} -3 & 1 & 2 & -9 & -18 \ & & -3 & 3 & 18 \ \hline & 1 & -1 & -6 & 0 \ \end{array} \]
The quotient is \(x^2 - x - 6\).
Next, we need to factor \(x^2 - x - 6\):
To factor \(x^2 - x - 6\), we look for two numbers that multiply to \(-6\) and add to \(-1\). The numbers \(-3\) and \(2\) work:
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
Now we can write the full factorization of the original polynomial:
\[ x^3 + 2x^2 - 9x - 18 = (x + 3)(x - 3)(x + 2) \]
Thus, the complete factorization of \(x^3 + 2x^2 - 9x - 18\) is:
\[ (x + 3)(x - 3)(x + 2) \]