Given the equation $$ x^{4}-6 x^{3}-5 x^{2}+24 x+4=0 $$, complete the following. a. List all possible rational roots. b. Use synthetic division to test several possible rational roots in order to identify one actual root. c. Use the root from part (b) to solve the equation. a. List all rational roots that are possible according to the Rational Zero Theorem. $$ \pm 1, \pm 2, \pm 4 $$ (Use commas to separate answers as needed.) b. Use synthetic division to test several possible rational roots in order to identify one actual root. One rational root of the given equation is (Simplify your answer.)

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To solve the equation $$ x^{4}-6 x^{3}-5 x^{2}+24 x+4=0 $$, we will follow the steps outlined in your request.

### Part a: List all possible rational roots.

According to the Rational Root Theorem, the possible rational roots of a polynomial are given by the factors of the constant term divided by the factors of the leading coefficient.

For the polynomial $$ x^{4}-6 x^{3}-5 x^{2}+24 x+4 $$:
- The constant term is $$ 4 $$.
- The leading coefficient is $$ 1 $$.

The factors of $$ 4 $$ are $$ \pm 1, \pm 2, \pm 4 $$. Since the leading coefficient is $$ 1 $$, the possible rational roots are:

$$
\pm 1, \pm 2, \pm 4
$$

### Part b: Use synthetic division to test several possible rational roots.

Let's test the possible rational roots using synthetic division. We will start with $$ x = 1 $$:

**Testing $$ x = 1 $$:**

$$
\begin{array}{r|rrrrr}
1 & 1 & -6 & -5 & 24 & 4 \
  &   & 1 & -5 & -10 & 14 \
\hline
  & 1 & -5 & -10 & 14 & 18 \
\end{array}
$$

The remainder is $$ 18 $$, so $$ x = 1 $$ is not a root.

**Testing $$ x = -1 $$:**

$$
\begin{array}{r|rrrrr}
-1 & 1 & -6 & -5 & 24 & 4 \
   &   & -1 & 7 & -2 & -22 \
\hline
   & 1 & -7 & 2 & 22 & -18 \
\end{array}
$$

The remainder is $$ -18 $$, so $$ x = -1 $$ is not a root.

**Testing $$ x = 2 $$:**

$$
\begin{array}{r|rrrrr}
2 & 1 & -6 & -5 & 24 & 4 \
  &   & 2 & -8 & -10 & 28 \
\hline
  & 1 & -4 & -13 & 14 & 32 \
\end{array}
$$

The remainder is $$ 32 $$, so $$ x = 2 $$ is not a root.

**Testing $$ x = -2 $$:**

$$
\begin{array}{r|rrrrr}
-2 & 1 & -6 & -5 & 24 & 4 \
   &   & -2 & 16 & -22 & -4 \
\hline
   & 1 & -8 & 11 & 2 & 0 \
\end{array}
$$

The remainder is $$ 0 $$, so $$ x = -2 $$ is a root.

### Part c: Use the root from part (b) to solve the equation.

Since we found that $$ x = -2 $$ is a root, we can factor the polynomial as follows:

$$
x^{4}-6 x^{3}-5 x^{2}+24 x+4 = (x + 2)(x^{3}-8 x^{2}+11 x + 2)
$$

Now we need to solve the cubic polynomial $$ x^{3}-8 x^{2}+11 x + 2 = 0 $$. We can use synthetic division again to find more roots or apply the Rational Root Theorem again.

**Testing $$ x = 1 $$ on the cubic:**

$$
\begin{array}{r|rrrr}
1 & 1 & -8 & 11 & 2 \
  &   & 1 & -7 & 4 \
\hline
  & 1 & -7 & 4 & 6 \
\end{array}
$$

The remainder is $$ 6 $$, so $$ x = 1 $$ is not a root.

**Testing $$ x = 2 $$ on the cubic:**

$$
\begin{array}{r|rrrr}
2 & 1 & -8 & 11 & 2 \
  &   & 2 & -12 & -2 \
\hline
  & 1 & -6 & -1 & 0 \
\end{array}
$$

The remainder is $$ 0 $$, so $$ x = 2 $$ is a root of the cubic polynomial.

Now we can factor the cubic polynomial as:

$$
x^{3}-8 x^{2}+11 x + 2 = (x - 2)(x^{2}-6 x - 1)
$$

Now we need to solve $$ x^{2}-6 x - 1 = 0 $$ using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 + 4}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10}
$$

### Final Roots

Thus, the roots of the original polynomial $$ x^{4}-6 x^{3}-5 x^{2}+24 x+4=0 $$ are:

$$
x = -2, \quad x = 2, \quad x = 3 + \sqrt{10}, \quad x = 3 - \sqrt{10}
$$

### Summary

- **Possible rational roots:** $$ \pm 1, \pm 2, \pm 4 $$
- **One actual root:** $$ -2 $$
a. $$ \pm 1, \pm 2, \pm 4 $$ b. $$ -2 $$

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Solución

Part a: List all possible rational roots.

Part b: Use synthetic division to test several possible rational roots.

Part c: Use the root from part (b) to solve the equation.

Final Roots

Summary

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