Question

find the zeros for the polynomial function and give the multiplicity for each zero. state whether the graph crosses the x - axis or touches the x - axis and turns around at each zero.
$f(x)=x^{3}+7x^{2}-16x - 112$
determine the zero(s), if they exist.
the zero(s) is/are
(type integers or decimals. use a comma to separate answers as needed.)
determine the multiplicities of the zero(s), if they exist. select the correct choice below and, if necessary, fill in the answer box(es) within your choice.
a. there are two zeros. the multiplicity of the smallest zero is. the multiplicity of the largest zero is
(simplify your answers.)
b. there is one zero. the multiplicity of the zero is
(simplify your answer.)
c. there are three zeros. the multiplicity of the smallest zero is. the multiplicity of the largest zero is. the multiplicity of the other zero is
(simplify your answers.)

Explanation:

Step1: Factor the polynomial

Use factoring by grouping:

$$\begin{align*} f(x)&=x^3 + 7x^2 - 16x - 112\\ &=x^2(x+7)-16(x+7)\\ &=(x+7)(x^2-16)\\ &=(x+7)(x-4)(x+4) \end{align*}$$

Step2: Find the zeros

Set $f(x)=0$:
$(x+7)(x-4)(x+4)=0$
Solve for $x$: $x=-7$, $x=4$, $x=-4$

Step3: Identify multiplicities

Each linear factor has an exponent of 1, so each zero has multiplicity 1.

Step4: Determine x-axis behavior

For a zero with odd multiplicity, the graph crosses the x-axis. Since all multiplicities are 1 (odd), the graph crosses the x-axis at each zero.

Answer:

The zeros are $-7, 4, -4$; Choice C: The multiplicity of the smallest zero $-7$ is $1$, the multiplicity of the largest zero $4$ is $1$, the multiplicity of the other zero $-4$ is $1$.
For the x-axis behavior: The graph crosses the x-axis at $x=-7$, $x=4$, and $x=-4$.