Question

find all the zeros of the polynomial function. use the rational zero theorem, descartess rule of signs, and possibly the graph of the polynomial function shown by a graphing utility as an aid in obtaining the first zero.
$f(x) = x^3 - 3x^2 - 6x + 8$
the zeros of the function are \boxed{}.
(use a comma to separate your answers as needed.)

Explanation:

Step1: Apply Rational Zero Theorem

The Rational Zero Theorem states that possible rational zeros are $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. For $f(x)=x^3 - 3x^2 - 6x + 8$, leading coefficient is $1$ (factors $q=\pm1$) and constant term is $8$ (factors $p=\pm1,\pm2,\pm4,\pm8$). So possible rational zeros: $\pm1,\pm2,\pm4,\pm8$.

Step2: Test possible zeros

Test $x = 1$: $f(1)=1 - 3 - 6 + 8 = 0$. So $x = 1$ is a zero.

Step3: Factor the polynomial

Since $x = 1$ is a zero, $(x - 1)$ is a factor. Use polynomial division or synthetic division. Using synthetic division with root $1$:

$$
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$$

So $f(x)=(x - 1)(x^2 - 2x - 8)$.

Step4: Factor the quadratic

Factor $x^2 - 2x - 8$: find two numbers that multiply to $-8$ and add to $-2$. These are $-4$ and $2$. So $x^2 - 2x - 8=(x - 4)(x + 2)$.

Step5: Find all zeros

Set each factor to zero: $x - 1 = 0 \Rightarrow x = 1$; $x - 4 = 0 \Rightarrow x = 4$; $x + 2 = 0 \Rightarrow x = -2$.

Answer:

-2, 1, 4