Questions: Consider the following function. g(x)=x^3+4x^2-4x-16 (a) Find all real zeros of the polynomial function. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) x= (b) Determine whether the multiplicity of each zero is even or odd. smaller x-value ---Select--- ---Select--- larger x-value ---Select--- (c) Determine the maximum possible number of turning points of the graph of the function. Use a graphing utility to graph the function and verify your answers. turning point(s)

Solution Steps

To solve the given polynomial function \( g(x) = x^3 + 4x^2 - 4x - 16 \), we need to: (a) Find all real zeros of the polynomial function. (b) Determine the multiplicity of each zero. (c) Determine the maximum possible number of turning points of the graph of the function.

1. Finding Real Zeros: Use numerical methods or a root-finding algorithm to find the real zeros of the polynomial.
2. Multiplicity of Zeros: Analyze the polynomial to determine the multiplicity of each zero.
3. Turning Points: The maximum number of turning points of a polynomial function is given by \( n-1 \), where \( n \) is the degree of the polynomial.

To find the real zeros of the polynomial function \( g(x) = x^3 + 4x^2 - 4x - 16 \), we need to solve the equation \( g(x) = 0 \).

First, we use the Rational Root Theorem to identify possible rational roots. The Rational Root Theorem states that any rational root, in the form of \( \frac{p}{q} \), must be a factor of the constant term (-16) divided by a factor of the leading coefficient (1).

Factors of -16: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \)

Factors of 1: \( \pm 1 \)

Possible rational roots: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \)

We test these possible roots by substituting them into the polynomial:

1. \( g(1) = 1^3 + 4(1)^2 - 4(1) - 16 = 1 + 4 - 4 - 16 = -15 \) (not a root)
2. \( g(-1) = (-1)^3 + 4(-1)^2 - 4(-1) - 16 = -1 + 4 + 4 - 16 = -9 \) (not a root)
3. \( g(2) = 2^3 + 4(2)^2 - 4(2) - 16 = 8 + 16 - 8 - 16 = 0 \) (root found)

Since \( x = 2 \) is a root, we can factor \( g(x) \) as \( (x - 2) \) times a quadratic polynomial. We perform polynomial division to find the quadratic factor:

\[ \begin{array}{r|rrr} 2 & 1 & 4 & -4 & -16 \\ & & 2 & 12 & 16 \\ \hline & 1 & 6 & 8 & 0 \\ \end{array} \]

So, \( g(x) = (x - 2)(x^2 + 6x + 8) \).

Next, we solve the quadratic equation \( x^2 + 6x + 8 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1 \), \( b = 6 \), and \( c = 8 \):

\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm \sqrt{4}}{2} = \frac{-6 \pm 2}{2} \]

Thus, the solutions are:

\[ x = \frac{-6 + 2}{2} = -2 \quad \text{and} \quad x = \frac{-6 - 2}{2} = -4 \]

So, the real zeros of the polynomial function are \( x = 2, -2, -4 \).

\[ \boxed{x = 2, -2, -4} \]

The polynomial \( g(x) = (x - 2)(x + 2)(x + 4) \) shows that each zero has a multiplicity of 1, which is odd.

• Smaller \( x \)-value: \( x = -4 \) (odd multiplicity)
• Larger \( x \)-value: \( x = 2 \) (odd multiplicity)

\[ \boxed{\text{odd}} \]

The maximum number of turning points of a polynomial function is given by \( n - 1 \), where \( n \) is the degree of the polynomial. For \( g(x) = x^3 + 4x^2 - 4x - 16 \), the degree \( n = 3 \).

Thus, the maximum possible number of turning points is:

\[ 3 - 1 = 2 \]

Using a graphing utility to graph the function confirms that there are indeed 2 turning points.

\[ \boxed{2 \text{ turning points}} \]

Final Answer