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}+6x^{2}-4x - 24
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: Group the terms

$f(x)=(x^{3}+6x^{2})-(4x + 24)$

Step2: Factor out common factors from each group

$x^{2}(x + 6)-4(x + 6)$

Step3: Factor out the common binomial factor

$f(x)=(x + 6)(x^{2}-4)$

Step4: Factor the difference - of - squares

$f(x)=(x + 6)(x + 2)(x - 2)$

Step5: Set $f(x)=0$ to find the zeros

$x+6=0$ gives $x=-6$; $x + 2=0$ gives $x=-2$; $x-2=0$ gives $x = 2$

The multiplicity of $x=-6$ is 1, the multiplicity of $x=-2$ is 1, and the multiplicity of $x = 2$ is 1. Since the multiplicity of each zero is 1 (an odd number), the graph of the function crosses the x - axis at $x=-6$, $x=-2$, and $x = 2$.

Answer:

$-6,-2,2$