To find the zeros of the polynomial function $f(x) = x^3 + 6x^2 - 4x - 24$, we'll follow these steps:

1. Factor the polynomial to find its zeros.
2. Determine the multiplicity of each zero.
3. Analyze the behavior of the graph at each zero (whether it crosses the x-axis or touches it and turns around).

Step 1: Factor the Polynomial

We first attempt to factor the cubic polynomial.

Since this is a cubic polynomial, we can start by checking if there is an obvious root (zero) by using the Rational Root Theorem, which suggests that any rational root must be a factor of the constant term ($-24$) divided by the leading coefficient ($1$). The possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.

We can test these by substituting them into the polynomial.

Let's test $x = 2$:

$f(2) = (2)^3 + 6(2)^2 - 4(2) - 24 = 8 + 24 - 8 - 24 = 0$

So, $x = 2$ is a root.

Now, we can factor $x - 2$ out of $f(x)$:

Using synthetic division:

2 & 1 & 6 & -4 & -24 \\ & & 2 & 16 & 24 \\ \hline & 1 & 8 & 12 & 0 \\ \end{array}$$ This gives us: $$f(x) = (x - 2)(x^2 + 8x + 12)$$ ### Step 2: Factor the Quadratic Polynomial Next, factor the quadratic $$x^2 + 8x + 12$$ if possible: The quadratic formula can be used here: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$a = 1$$, $$b = 8$$, and $$c = 12$$: $$x = \frac{-8 \pm \sqrt{64 - 48}}{2} = \frac{-8 \pm \sqrt{16}}{2} = \frac{-8 \pm 4}{2}$$ So: $$x = \frac{-8 + 4}{2} = -2 \quad \text{or} \quad x = \frac{-8 - 4}{2} = -6$$ Thus, the quadratic factors as: $$x^2 + 8x + 12 = (x + 2)(x + 6)$$ ### Step 3: Write the Fully Factored Form The fully factored form of $$f(x)$$ is: $$f(x) = (x - 2)(x + 2)(x + 6)$$ ### Step 4: Determine Zeros and Their Multiplicity The zeros of the function are: - $$x = 2$$ with multiplicity 1. - $$x = -2$$ with multiplicity 1. - $$x = -6$$ with multiplicity 1. ### Step 5: Analyze the Behavior at Each Zero - **At $$x = 2$$**: The zero has an odd multiplicity (1), so the graph crosses the x-axis at $$x = 2$$. - **At $$x = -2$$**: The zero has an odd multiplicity (1), so the graph crosses the x-axis at $$x = -2$$. - **At $$x = -6$$**: The zero has an odd multiplicity (1), so the graph crosses the x-axis at $$x = -6$$. ### Final Answer: - The zeros are $$x = 2$$, $$x = -2$$, and $$x = -6$$, each with a multiplicity of 1. - The graph crosses the x-axis at each zero. Would you like further details or have any other questions? Here are five related questions you might find interesting: 1. How do you use the Rational Root Theorem to find potential zeros? 2. What is the significance of multiplicity in determining the behavior of a graph? 3. Can a polynomial of degree 3 have fewer than three real zeros? 4. How does the shape of the graph of a cubic function generally look like? 5. How does synthetic division work, and when is it used? **Tip:** Always check for common factors first when factoring polynomials, as it can simplify the process significantly.