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 + 8x^2 - 4x - 32 determine the zero(s), if they exist. the zero(s) is/are . (type integers or decimals. use a comma to separate answers as needed.)

Explanation:

Step1: Factor the polynomial by grouping

Group the terms: $(x^{3}+8x^{2})+(-4x - 32)$.
Factor out the greatest - common factor from each group: $x^{2}(x + 8)-4(x + 8)$.
Then factor out $(x + 8)$: $f(x)=(x + 8)(x^{2}-4)$.

Step2: Factor the difference of squares

Since $x^{2}-4$ is a difference of squares ($a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 2$), we have $x^{2}-4=(x + 2)(x - 2)$.
So $f(x)=(x + 8)(x + 2)(x - 2)$.

Step3: Find the zeros

Set $f(x)=0$. Then $(x + 8)(x + 2)(x - 2)=0$.
Using the zero - product property (if $ab = 0$, then $a = 0$ or $b = 0$), we get:
$x+8=0\Rightarrow x=-8$; $x + 2=0\Rightarrow x=-2$; $x - 2=0\Rightarrow x = 2$.
The multiplicity of each zero is 1 because each factor $(x + 8),(x + 2),(x - 2)$ appears once.
Since the multiplicity of each zero is 1 (an odd number), the graph of the function crosses the x - axis at each zero.

Answer:

$-8,-2,2$