26. Which binomial IS afactor of x^3+6x^2+9x+54 (x-6) (x-3) (x^2+9) (x+3)

To determine which binomial is a factor of the polynomial $x^3 + 6x^2 + 9x + 54$, we can use the Factor Theorem. The Factor Theorem states that if $x = a$ is a root of the polynomial, then $(x - a)$ is a factor of the polynomial.

Let's test each binomial to see if it is a factor:

1. **Test $(x - 6)$:**
Substitute $x = 6$ into the polynomial:
$$6^3 + 6 \cdot 6^2 + 9 \cdot 6 + 54 = 216 + 216 + 54 + 54 = 540 \neq 0$$
So, $(x - 6)$ is not a factor.

2. **Test $(x - 3)$:**
Substitute $x = 3$ into the polynomial:
$$3^3 + 6 \cdot 3^2 + 9 \cdot 3 + 54 = 27 + 54 + 27 + 54 = 162 \neq 0$$
So, $(x - 3)$ is not a factor.

3. **Test $(x^2 + 9)$:**
Substitute $x = 0$ into the polynomial:
$$0^3 + 6 \cdot 0^2 + 9 \cdot 0 + 54 = 54 \neq 0$$
So, $(x^2 + 9)$ is not a factor.

4. **Test $(x + 3)$:**
Substitute $x = -3$ into the polynomial:
$$(-3)^3 + 6 \cdot (-3)^2 + 9 \cdot (-3) + 54 = -27 + 54 - 27 + 54 = 54 \neq 0$$
So, $(x + 3)$ is not a factor.

Since none of the substitutions resulted in zero, it appears there was an error in the initial polynomial factorization. Let's re-evaluate the polynomial by factoring it directly.

We can try factoring by grouping:
$$x^3 + 6x^2 + 9x + 54$$

Group the terms:
$$(x^3 + 6x^2) + (9x + 54)$$

Factor out the common terms:
$$x^2(x + 6) + 9(x + 6)$$

Factor out the common binomial factor $(x + 6)$:
$$(x^2 + 9)(x + 6)$$

So, the factors of the polynomial are $(x^2 + 9)$ and $(x + 6)$.

Therefore, the correct answer is:
$$(x^2 + 9)$$