To factor the polynomial [tex]\( x^3 + 6x^2 - 9x - 54 \)[/tex] completely, we can follow these steps:
1. Identify a possible rational root:
According to the Rational Root Theorem, any rational root of the polynomial [tex]\( ax^n + bx^{n-1} + \cdots + c \)[/tex] is a factor of the constant term (in this case, -54) divided by a factor of the leading coefficient (in this case, 1). Therefore, possible rational roots are [tex]\( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 27, \pm 54 \)[/tex].
2. Test the potential roots:
Let's test [tex]\( x = -3 \)[/tex]:
Substitute [tex]\( x = -3 \)[/tex] into the polynomial:
[tex]\[
(-3)^3 + 6(-3)^2 - 9(-3) - 54 = -27 + 54 + 27 - 54 = 0
\][/tex]
So, [tex]\( x = -3 \)[/tex] is a root.
3. Perform polynomial division:
Since [tex]\( x = -3 \)[/tex] is a root, [tex]\( x + 3 \)[/tex] is a factor. Next, we divide [tex]\( x^3 + 6x^2 - 9x - 54 \)[/tex] by [tex]\( x + 3 \)[/tex].
Using synthetic division or long division:
[tex]\[
\begin{array}{r|rrrr}
-3 & 1 & 6 & -9 & -54 \\
& & -3 & -9 & 54 \\
\hline
& 1 & 3 & -18 & 0 \\
\end{array}
\][/tex]
The quotient is [tex]\( x^2 + 3x - 18 \)[/tex], and the remainder is 0.
4. Factor the quotient quadratic polynomial:
Now we need to factor [tex]\( x^2 + 3x - 18 \)[/tex].
We look for two numbers that multiply to [tex]\(-18\)[/tex] and add to [tex]\(3\)[/tex]. These numbers are [tex]\(6\)[/tex] and [tex]\(-3\)[/tex].
So, we can factor [tex]\( x^2 + 3x - 18 \)[/tex] as:
[tex]\[
x^2 + 3x - 18 = (x + 6)(x - 3)
\][/tex]
5. Combine the factors:
Therefore, the complete factorization of the original polynomial is:
[tex]\[
x^3 + 6x^2 - 9x - 54 = (x + 3)(x + 6)(x - 3)
\][/tex]
Thus, the correct factorization is:
[tex]\[
(x+3)(x-3)(x+6)
\][/tex]
The correct answer is:
[tex]\[
(x + 3)(x - 3)(x + 6)
\][/tex]
