You provided the polynomial:

$P(x) = x^3 + x^2 - 2x - 8$

Step 1: Check for Rational Roots

Using the Rational Root Theorem, the possible rational roots are the factors of $-8$ (constant term) divided by the factors of $1$ (leading coefficient):

$\pm 1, \pm 2, \pm 4, \pm 8$

Let's test $x = 2$:

$(2)^3 + (2)^2 - 2(2) - 8 = 8 + 4 - 4 - 8 = 0$

Since $x = 2$ is a root, we can factor $x - 2$ from $P(x)$.

Step 2: Polynomial Division

Dividing $P(x)$ by $(x - 2)$ using synthetic division:

  2 |  1   1  -2  -8  
    |      2   6   8  
  ------------------
      1   3   4   0  

The quotient is:

$x^2 + 3x + 4$

Step 3: Solve $x^2 + 3x + 4 = 0$

Using the quadratic formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(4)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 - 16}}{2}$

$x = \frac{-3 \pm \sqrt{-7}}{2}$

Since $\sqrt{-7}$ is imaginary, the remaining roots are:

$x = \frac{-3 \pm i\sqrt{7}}{2}$

Final Answer:

The factorization of $P(x)$:

$(x - 2) \left(x^2 + 3x + 4\right)$

The roots:

$x = 2, \quad x = \frac{-3 + i\sqrt{7}}{2}, \quad x = \frac{-3 - i\sqrt{7}}{2}$

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Would you like more details or have any questions?

Related Questions:

1. How does the Rational Root Theorem help in solving polynomials?
2. What are the steps for performing synthetic division?
3. How do you solve quadratic equations with complex roots?
4. Can this polynomial be solved using another method, such as factoring by grouping?
5. How do complex roots always appear in conjugate pairs?

Tip: Always check for rational roots first using the Rational Root Theorem—it simplifies polynomial solving significantly!