Question

Find all the zeros of the polynomial function. Use the Rational Zero​ Theorem, Descartes's 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 left parenthesis x right parenthesis equals x cubed minus 3 x squared minus 6 x plus 8
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Part 1
The zeros of the function are
  
enter your response here.
​(Use a comma to separate your answers as​ needed.)

Answer 1

To find the zeros of the polynomial function \( f(x) = x^3 - 3x^2 - 6x + 8 \), we can apply the Rational Root Theorem and Descartes' Rule of Signs, alongside potentially using a graphing utility.

### Step 1: Use the Rational Root Theorem

According to the Rational Root Theorem, any rational solution (root) of the polynomial function is of the form \( \frac{p}{q} \), where:
- \( p \) is a factor of the constant term (which is 8), and
- \( q \) is a factor of the leading coefficient (which is 1).

The factors of 8 are ±1, ±2, ±4, ±8, so the possible rational roots are:

\[
\pm 1, \pm 2, \pm 4, \pm 8
\]

### Step 2: Use Descartes' Rule of Signs

Now let's apply Descartes' Rule of Signs:

- For \( f(x) \):
- The sign changes are as follows:
- \( +1 \) (for \( x^3 \)) to \( -3 \) (for \( -3x^2 \)) → 1 sign change
- \( -3 \) (for \( -3x^2 \)) to \( -6 \) (for \( -6x \)) → 0 sign changes
- \( -6 \) (for \( -6x \)) to \( +8 \) (for \( +8 \)) → 1 sign change
- Total sign changes = 2 (1 from \( + \) to \( - \) and 1 from \( - \) to \( + \))

So, there are either 2 or 0 positive real roots.

- For \( f(-x) \):
\[
f(-x) = -x^3 - 3(-x)^2 - 6(-x) + 8 = -x^3 - 3x^2 + 6x + 8
\]
The sign changes are:
- \( -1 \) (for \( -x^3 \)) to \( -3 \) (for \( -3x^2 \)) → 0 sign changes
- \( -3 \) (for \( -3x^2 \)) to \( +6 \) (for \( +6x \)) → 1 sign change
- \( +6 \) (for \( +6x \)) to \( +8 \) (for \( +8 \)) → 0 sign changes
- Total sign changes = 1

So, there is 1 negative real root.

### Step 3: Find the First Zero

Now we can use a graphing utility to find a possible root or we can test the rational roots we found. Testing the possible roots:

- Test \( x = 1 \):
\[
f(1) = 1^3 - 3(1^2) - 6(1) + 8 = 1 - 3 - 6 + 8 = 0
\]
So, \( x = 1 \) is a root.

### Step 4: Factor the Polynomial

Next, we can factor out \( (x - 1) \) from \( f(x) \) using synthetic division:

\[
\begin{array}{r|rrrr}
1 & 1 & -3 & -6 & 8 \\
& & 1 & -2 & -8 \\
\hline
& 1 & -2 & -8 & 0 \\
\end{array}
\]
The resulting polynomial is \( x^2 - 2x - 8 \).

### Step 5: Factor the Quadratic

Now we solve \( x^2 - 2x - 8 = 0 \) either by factoring or using the quadratic formula:

\[
x^2 - 2x - 8 = (x - 4)(x + 2) = 0
\]

### Step 6: Solve for Zeros

Setting each factor equal to zero gives us the solutions:

\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]

### Final Answer

Thus, the zeros of the polynomial \( f(x) = x^3 - 3x^2 - 6x + 8 \) are:

\[
\boxed{1, 4, -2}
\]