We are asked to factor the polynomial $4x^3 - 8x^2 - 36x - 72$.

1. Problem Description

We are asked to factor the polynomial $4x^3 - 8x^2 - 36x - 72$.

2. Solution Steps

First, we look for a common factor in all terms. We see that each term is divisible by

4. So we factor out 4:

$4x^3 - 8x^2 - 36x - 72 = 4(x^3 - 2x^2 - 9x - 18)$

Now, we try to factor the cubic polynomial $x^3 - 2x^2 - 9x - 18$ by grouping.

Group the first two terms and the last two terms:

$x^3 - 2x^2 - 9x - 18 = (x^3 - 2x^2) + (-9x - 18)$

Factor out the greatest common factor from each group. From the first group, we can factor out $x^2$. From the second group, we can factor out $-9$:

$(x^3 - 2x^2) + (-9x - 18) = x^2(x - 2) - 9(x + 2)$

There is an error in the problem. The problem should be $4x^3 - 8x^2 - 36x + 72$.

Then it will become $4(x^3-2x^2-9x+18) = 4((x^3-2x^2) + (-9x+18)) = 4(x^2(x-2)-9(x-2)) = 4(x-2)(x^2-9)$.

Since $x^2-9$ can be factored as the difference of squares, $x^2-9 = (x-3)(x+3)$.

Therefore $4x^3 - 8x^2 - 36x + 72 = 4(x-2)(x-3)(x+3)$.

Let's assume the problem meant $4x^3 - 8x^2 - 36x + 72$.

$4(x^3 - 2x^2 - 9x + 18)$

$4[x^2(x - 2) - 9(x - 2)]$

$4(x - 2)(x^2 - 9)$

$4(x - 2)(x - 3)(x + 3)$

3. Final Answer

Assuming there was a typo and the problem was $4x^3 - 8x^2 - 36x + 72$, then the factored form is $4(x-2)(x-3)(x+3)$.

Otherwise, it cannot be easily factored with integers. We factored out 4, so $4(x^3 - 2x^2 - 9x - 18)$. Factoring by grouping $x^2(x-2) - 9(x+2)$.

Final Answer: Assuming the problem was $4x^3 - 8x^2 - 36x + 72$, the answer is $4(x-3)(x+3)(x-2)$.

If the problem is as written, then $4(x^3 - 2x^2 - 9x - 18)$, and we cannot factor further using simple methods.

Since the question probably has a sign error, $4(x-3)(x+3)(x-2)$ is likely the answer.