Let's break down the function from the image:

$g(x) = -5x \left( x^2 - 1 \right) \left( x^2 + 36 \right)$

We need to find the real zeros of the function. Let's analyze it step-by-step:

Step 1: Setting $g(x) = 0$

We set the function equal to 0:

$-5x \left( x^2 - 1 \right) \left( x^2 + 36 \right) = 0$

Since this is a product of multiple terms, the only way the product is 0 is if any of the terms equals 0.

Step 2: Solve for Each Factor

1. First factor: $-5x = 0 \quad \Rightarrow \quad x = 0$

2. Second factor: $x^2 - 1 = 0$ $x^2 = 1 \quad \Rightarrow \quad x = \pm 1$

3. Third factor: $x^2 + 36 = 0$ $x^2 = -36$ This equation has no real solutions, since the square of a real number cannot be negative.

Step 3: List the Real Zeros

Thus, the only real solutions are from the first two factors:

$x = 0, \quad x = 1, \quad x = -1$

Final Answer

The real zeros are:

$\boxed{0, 1, -1}$

Let me know if you need further clarification or have any other questions! Here are a few related questions:

1. How do imaginary numbers affect finding roots?
2. What is the difference between real and complex solutions?
3. How would the graph of this function behave at these zeros?
4. How does multiplicity affect the behavior at a zero?
5. What if the polynomial had an extra $x$-term in the factor $(x^2 + 36)$?

Tip: Whenever dealing with factorizations, carefully check for both real and complex roots to avoid missing any solutions.