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}+4x^{2}-9x - 36
the zeros are
(use a comma to separate answers as needed.)

Explanation:

Step1: Factor by grouping

Group the terms: $f(x)=(x^{3}+4x^{2})-(9x + 36)$.
Factor out common factors from each group: $x^{2}(x + 4)-9(x + 4)=(x + 4)(x^{2}-9)$.

Step2: Factor the difference - of - squares

Since $x^{2}-9=(x + 3)(x - 3)$ (using the formula $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 3$), then $f(x)=(x + 4)(x + 3)(x - 3)$.

Step3: Find the zeros

Set $f(x)=0$.
If $(x + 4)(x + 3)(x - 3)=0$, then by the zero - product property, $x+4 = 0$ gives $x=-4$, $x + 3=0$ gives $x=-3$ and $x - 3=0$ gives $x = 3$.
The multiplicity of each zero is 1 because each factor $(x + 4),(x + 3),(x - 3)$ has an exponent of 1.
Since the multiplicity of each zero is 1 (an odd number), the graph of the function crosses the x - axis at $x=-4,x=-3$ and $x = 3$.

Answer:

$-4,-3,3$