To factor the polynomial \( x^3 + 2x^2 - 9x - 18 \), we can use the factor theorem and trial and error for possible rational roots.
Let's begin by finding a possible rational root using the Rational Root Theorem. We can try \( x = -3 \) since it is a factor of the constant term \(-18\):
\[ f(-3) = (-3)^3 + 2(-3)^2 - 9(-3) - 18 \] \[ = -27 + 18 + 27 - 18 \] \[ = 0 \]
Since \( f(-3) = 0 \), \( x + 3 \) is a factor. Now we can factor \( x^3 + 2x^2 - 9x - 18 \) by performing synthetic division:
\[ \begin{array}{r|rrrr} -3 & 1 & 2 & -9 & -18 \ & & -3 & 3 & 18 \ \hline & 1 & -1 & -6 & 0 \ \end{array} \]
The result of the synthetic division is \( x^2 - x - 6 \).
Now, we factor \( x^2 - x - 6 \):
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
Putting it all together, we have:
\[ x^3 + 2x^2 - 9x - 18 = (x + 3)(x - 3)(x + 2) \]
Thus, the complete factorization is: \[ (x + 3)(x - 3)(x + 2) \]
So, the missing pieces you can fill in are:
- First factor: \( (x + 3) \)
- Second factor: \( (x - 3) \)
- Third factor: \( (x + 2) \)