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 \boxed{}.
(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.
\bigcirc a. there are two zeros. the multiplicity of the smallest zero is \boxed{}. the multiplicity of the largest zero is \boxed{}.
(simplify your answers.)
\bigcirc b. there is one zero. the multiplicity of the zero is \boxed{}.
(simplify your answer.)
\bigcirc c. there are three zeros. the multiplicity of the smallest zero is \boxed{}. the multiplicity of the largest zero is \boxed{}. the multiplicity of the other zero is \boxed{}.
(simplify your answers.)

Explanation:

Step1: Factor by grouping

Group terms to factor:
$$f(x) = (x^3 + 7x^2) + (-16x - 112)$$
$$f(x) = x^2(x + 7) - 16(x + 7)$$
$$f(x) = (x + 7)(x^2 - 16)$$

Step2: Factor difference of squares

Factor $x^2-16$ as a difference of squares:
$$f(x) = (x + 7)(x - 4)(x + 4)$$

Step3: Find zeros

Set each factor equal to 0:
$x + 7 = 0 \implies x = -7$
$x - 4 = 0 \implies x = 4$
$x + 4 = 0 \implies x = -4$

Step4: Identify multiplicities

Each linear factor has degree 1, so each zero has multiplicity 1. For odd multiplicity, the graph crosses the x-axis.

Answer:

The zero(s) is/are $\boldsymbol{-7, -4, 4}$

$\boldsymbol{C. There are three zeros. The multiplicity of the smallest zero is \boldsymbol{1}. The multiplicity of the largest zero is \boldsymbol{1}. The multiplicity of the other zero is \boldsymbol{1}.}$

At $x=-7$: the graph crosses the x-axis.
At $x=-4$: the graph crosses the x-axis.
At $x=4$: the graph crosses the x-axis.