Find all the real zeros of the polynomial. Separate answers by commas. Enter exact answers using integers and fractions, not decimal approximations.  f ( x ) =     x^4 + 3x^3 − 6x^2 − 28x − 24    

To find the real zeros of the polynomial \( f(x) = x^4 + 3x^3 - 6x^2 - 28x - 24 \), we will use a combination of techniques such as the Rational Root Theorem, synthetic division, and factoring.

### Step 1: Use the Rational Root Theorem
The Rational Root Theorem suggests that any rational root of the polynomial must be a factor of the constant term (-24) divided by a factor of the leading coefficient (1). The factors of -24 are:

\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24
\]

We will test these possible roots by substituting them into the polynomial.

### Step 2: Test possible rational roots
We start by testing \( x = -2 \) as a possible root by substituting it into \( f(x) \).

\[
f(-2) = (-2)^4 + 3(-2)^3 - 6(-2)^2 - 28(-2) - 24 = 16 - 24 - 24 + 56 - 24 = 0
\]

Thus, \( x = -2 \) is a root.

### Step 3: Perform synthetic division
Now that we know \( x = -2 \) is a root, we can perform synthetic division of \( f(x) \) by \( x + 2 \) to reduce the polynomial.

We divide \( x^4 + 3x^3 - 6x^2 - 28x - 24 \) by \( x + 2 \).

\[
\begin{array}{r|rrrrr}
-2 & 1 & 3 & -6 & -28 & -24 \\
   &   & -2 & -2 & 16 & 24 \\
\hline
   & 1 & 1  & -8 & -12 & 0 \\
\end{array}
\]

The quotient is \( x^3 + x^2 - 8x - 12 \).

### Step 4: Factor the resulting cubic polynomial
Now, we need to find the zeros of \( x^3 + x^2 - 8x - 12 \). We again apply the Rational Root Theorem and test possible rational roots.

Testing \( x = 2 \):

\[
f(2) = 2^3 + 2^2 - 8(2) - 12 = 8 + 4 - 16 - 12 = -16 \quad \text{(not a root)}
\]

Testing \( x = -3 \):

\[
f(-3) = (-3)^3 + (-3)^2 - 8(-3) - 12 = -27 + 9 + 24 - 12 = -6 \quad \text{(not a root)}
\]

Testing \( x = 3 \):

\[
f(3) = 3^3 + 3^2 - 8(3) - 12 = 27 + 9 - 24 - 12 = 0
\]

Thus, \( x = 3 \) is a root.

### Step 5: Perform synthetic division again
We now perform synthetic division of \( x^3 + x^2 - 8x - 12 \) by \( x - 3 \).

\[
\begin{array}{r|rrrr}
3 & 1 & 1 & -8 & -12 \\
  &   & 3 & 12 & 12 \\
\hline
  & 1 & 4 & 4 & 0 \\
\end{array}
\]

The quotient is \( x^2 + 4x + 4 \).

### Step 6: Solve the quadratic equation
Now, we solve \( x^2 + 4x + 4 = 0 \) by factoring:

\[
x^2 + 4x + 4 = (x + 2)^2
\]

Thus, \( x = -2 \) is a double root.

### Step 7: Conclusion
The real zeros of the polynomial are:

\[
x = -2, -2, 3
\]

Thus, the real zeros are \( -2 \) (with multiplicity 2) and \( 3 \).

### Final Answer:
The real zeros are \( -2, -2, 3 \).

Would you like more details on any of the steps? 

Here are some follow-up questions you might explore:

1. What is the multiplicity of a root, and why is it important?
2. How does the Rational Root Theorem work in general?
3. Can synthetic division be used for non-rational roots?
4. How would the solution change if there were no rational roots?
5. What other methods can be used to solve higher-degree polynomials?

**Tip:** Always check your solutions by substituting the roots back into the original equation.

Find all the real zeros of the polynomial. Separate answers by commas. Enter exact answers using integers and fractions, not decimal approximations. f(x) = x^4 + 3x^3 − 6x^2 − 28x − 24

Discover how to find the real zeros of the polynomial f(x) = x^4 + 3x^3 - 6x^2 - 28x - 24 using techniques like the Rational Root Theorem and synthetic division. This problem highlights advanced polynomial factoring methods and is suitable for high school students in Grades 10-12.

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