, 6. $$ x^{3}+2 x^{2}-5 x-6=0 $$

Question

UpStudy AI · Accepted Answer

To solve the cubic equation:

$$
x^{3} + 2x^{2} - 5x - 6 = 0
$$

**Step 1: Find Rational Roots**

Using the Rational Root Theorem, possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. Here, the constant term is $$-6$$ and the leading coefficient is $$1$$. Therefore, the possible rational roots are:

$$
\pm1, \pm2, \pm3, \pm6
$$

**Step 2: Test Possible Roots**

Let's test these potential roots by substituting them into the equation.

- **Testing $$x = 1$$:**
  $$
  1^3 + 2(1)^2 - 5(1) - 6 = 1 + 2 - 5 - 6 = -8 \neq 0
  $$
  
- **Testing $$x = -1$$:**
  $$
  (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0
  $$
  
Since $$x = -1$$ is a root, we can factor $$(x + 1)$$ from the polynomial.

**Step 3: Factor the Polynomial**

Using synthetic division to factor out $$(x + 1)$$:

$$
\begin{array}{r|rrrr}
-1 & 1 & 2 & -5 & -6 \
   &   & -1 & -1 & 6 \
\hline
   & 1 & 1 & -6 & 0 \
\end{array}
$$

This leaves us with the quadratic equation:

$$
x^2 + x - 6 = 0
$$

**Step 4: Solve the Quadratic Equation**

Use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For $$x^2 + x - 6 = 0$$, $$a = 1$$, $$b = 1$$, and $$c = -6$$:

$$
x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm \sqrt{25}}{2} = \frac{-1 \pm 5}{2}
$$

This gives two solutions:

$$
x = \frac{-1 + 5}{2} = 2 \quad \text{and} \quad x = \frac{-1 - 5}{2} = -3
$$

**Step 5: Combine All Roots**

The solutions to the original cubic equation are:

$$
x = -3, \quad x = -1, \quad \text{and} \quad x = 2
$$

**Final Answer:**
$$
x = -3,\ -1,\ 2
$$
The solutions to the equation $$ x^{3} + 2x^{2} - 5x - 6 = 0 $$ are $$ x = -3 $$, $$ x = -1 $$, and $$ x = 2 $$.

, 6. \( x^{3}+2 x^{2}-5 x-6=0 \)

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