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: Group the terms

$f(x)=(x^{3}+8x^{2})-(4x + 32)$

Step2: Factor out common factors from each group

$=x^{2}(x + 8)-4(x + 8)$

Step3: Factor out the common binomial factor

$=(x + 8)(x^{2}-4)$

Step4: Factor the difference - of - squares

$=(x + 8)(x + 2)(x - 2)$

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

$x+8=0$ gives $x=-8$; $x + 2=0$ gives $x=-2$; $x - 2=0$ gives $x=2$

The multiplicity of each zero is 1 since the factors are of the first degree. When the multiplicity of a zero is 1, the graph of the function crosses the x - axis at that zero.

Answer:

-8,-2,2