Question

Fill in the missing pieces in the factorization of x3+2x2−9x−18
.

(1 point)

Answer 1

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: \( (x + 2) \)