To solve this problem, we are given a polynomial $h(x) = x^3 - x^2 - 24x - 36$ and the binomial $x + 2$. We are tasked with:

1. Showing that $x + 2$ is a factor of the polynomial.
2. Factoring the polynomial completely.

Step 1: Show that $x + 2$ is a factor of the polynomial

To check if $x + 2$ is a factor, we can use the Remainder Theorem, which states that if you substitute $x = -2$ into the polynomial and the result is zero, then $x + 2$ is a factor.

Let's substitute $x = -2$ into $h(x)$:

$h(-2) = (-2)^3 - (-2)^2 - 24(-2) - 36$

Now calculate:

$h(-2) = -8 - 4 + 48 - 36 = 0$

Since $h(-2) = 0$, the binomial $x + 2$ is indeed a factor of the polynomial $h(x)$.

Step 2: Factor the polynomial completely

Now, we will use synthetic division to divide $h(x)$ by $x + 2$ to find the quotient.

Synthetic Division:

We are dividing $x^3 - x^2 - 24x - 36$ by $x + 2$ using synthetic division.

1. Write the coefficients of $h(x)$: $[1, -1, -24, -36]$.
2. Perform synthetic division using $-2$ (since we're dividing by $x + 2$).

Steps:

• Bring down the first coefficient (1).
• Multiply $1 \times -2 = -2$, then add to the second coefficient: $-1 + (-2) = -3$.
• Multiply $-3 \times -2 = 6$, then add to the next coefficient: $-24 + 6 = -18$.
• Multiply $-18 \times -2 = 36$, then add to the next coefficient: $-36 + 36 = 0$.

Thus, the quotient is $x^2 - 3x - 18$, and the remainder is 0 (as expected).

So, the factorization so far is:

$h(x) = (x + 2)(x^2 - 3x - 18)$

Step 3: Factor the quadratic expression

Now we factor the quadratic $x^2 - 3x - 18$.

We need two numbers that multiply to $-18$ and add to $-3$. These numbers are $-6$ and $3$.

So, we can factor $x^2 - 3x - 18$ as:

$x^2 - 3x - 18 = (x - 6)(x + 3)$

Final Factorization

Thus, the complete factorization of $h(x)$ is:

$h(x) = (x + 2)(x - 6)(x + 3)$

Summary:

• We showed that $x + 2$ is a factor of the polynomial.
• The complete factorization is $h(x) = (x + 2)(x - 6)(x + 3)$.

Would you like more details on any of these steps or additional clarification?

Here are some questions that expand upon this explanation:

1. How does synthetic division work in general for polynomials?
2. Why does the Remainder Theorem confirm that $x + 2$ is a factor when the remainder is zero?
3. Could we use another method to verify the factorization, like the Factor Theorem or long division?
4. How would the factorization change if the binomial was different, for example, $x - 2$?
5. Can all cubic polynomials be factored in this way, or are there cases where they can't be factored over the real numbers?

Tip: Always check the roots of the polynomial after factoring to ensure they satisfy the original equation, as this confirms the correctness of your factorization.