To factor the cubic polynomial \( x^3 + 2x^2 - 9x - 18 \), we can use polynomial long division or trial and error to find one root.
Let's try to find a rational root using the Rational Root Theorem. We will test possible factors of \(-18\) (the constant term) over the factors of \(1\) (the leading coefficient):
Possible rational roots: \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
Testing \( x = -2 \):
\[ (-2)^3 + 2(-2)^2 - 9(-2) - 18 = -8 + 8 + 18 - 18 = 0. \]
So \( x = -2 \) is a root.
Now we can perform polynomial long division, dividing \( x^3 + 2x^2 - 9x - 18 \) by \( x + 2 \):
- Divide \( x^3 \) by \( x \) to get \( x^2 \).
- Multiply \( x^2 \) by \( x + 2 \) to get \( x^3 + 2x^2 \).
- Subtract \( (x^3 + 2x^2) \) from \( (x^3 + 2x^2 - 9x - 18) \) to get \( -9x - 18 \).
- Next, divide \( -9x \) by \( x \) to get \(-9\).
- Multiply \(-9\) by \( x + 2\) to get \(-9x - 18\).
- Subtract \( -9x - 18 \) from \(-9x - 18\) to get \(0\).
So, we can write:
\[ x^3 + 2x^2 - 9x - 18 = (x + 2)(x^2 - 9). \]
Now, we can factor \( x^2 - 9 \) further:
\[ x^2 - 9 = (x - 3)(x + 3). \]
Therefore, the entire factorization of \( x^3 + 2x^2 - 9x - 18 \) is:
\[ (x + 2)(x - 3)(x + 3). \]
For your original format \((x^2 - ?)(x + ?)\):
We can use \( x^2 - 9 \) as \( x^2 - 9 \) (where \( ? = 9 \)) and keep \( x + 2 \). Therefore, we can rewrite it as:
\[ (x^2 - 9)(x + 2). \]
So, the missing pieces in the factorization \((x^2 - ?)(x + ?)\) are:
? = 9 and ? = 2.
Thus, \((x^2 - 9)(x + 2)\) is the complete factorization.