To solve the cubic equation \( x^3 + 3x^2 - 4x - 12 = 0 \), we can use various methods such as factoring by grouping, synthetic division, or finding rational roots. Let's start by checking for rational roots using the Rational Root Theorem.

### Rational Root Theorem
The Rational Root Theorem states that any rational solution of the polynomial equation \( a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0 \) is a fraction \(\frac{p}{q}\), where \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \).

For the polynomial \( x^3 + 3x^2 - 4x - 12 \):

- The constant term \( a_0 = -12 \), so the factors of \(-12\) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
- The leading coefficient \( a_n = 1 \), so the factors of \( 1 \) are \( \pm 1 \).

Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

### Testing Possible Rational Roots

Let's test these possible roots by substituting them into the polynomial equation \( x^3 + 3x^2 - 4x - 12 \):

1. \( x = 1 \):
   \[
   1^3 + 3(1)^2 - 4(1) - 12 = 1 + 3 - 4 - 12 = -12 \neq 0
   \]
   So, \( x = 1 \) is not a root.

2. \( x = -1 \):
   \[
   (-1)^3 + 3(-1)^2 - 4(-1) - 12 = -1 + 3 + 4 - 12 = -6 \neq 0
   \]
   So, \( x = -1 \) is not a root.

3. \( x = 2 \):
   \[
   2^3 + 3(2)^2 - 4(2) - 12 = 8 + 12 - 8 - 12 = 0
   \]
   So, \( x = 2 \) is a root.

Since \( x = 2 \) is a root, we can factor \( x - 2 \) out of the polynomial.

### Factoring \( x^3 + 3x^2 - 4x - 12 \)

Divide the polynomial \( x^3 + 3x^2 - 4x - 12 \) by \( x - 2 \) using synthetic division or polynomial division.

#### Synthetic Division
\[
\begin{array}{r|rrrr}
2 & 1 & 3 & -4 & -12 \\
  &   & 2 & 10 & 12 \\
\hline
  & 1 & 5 & 6 & 0 \\
\end{array}
\]

The quotient is \( x^2 + 5x + 6 \).

So, \( x^3 + 3x^2 - 4x - 12 = (x - 2)(x^2 + 5x + 6) \).

### Solving \( x^2 + 5x + 6 = 0 \)
Factor \( x^2 + 5x + 6 \):
\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]

### Complete Factorization
\[
x^3 + 3x^2 - 4x - 12 = (x - 2)(x + 2)(x + 3)
\]

### Finding the Roots
Setting each factor to zero:
\[
x - 2 = 0 \implies x = 2
\]
\[
x + 2 = 0 \implies x = -2
\]
\[
x + 3 = 0 \implies x = -3
\]

Thus, the roots of the equation \( x^3 + 3x^2 - 4x - 12 = 0 \) are:
\[
x = 2, -2, -3
\]

Would you like any more details or have any questions?

### Related Questions
1. How can we verify the roots of a polynomial equation?
2. What is the Rational Root Theorem and how is it used?
3. Can you explain synthetic division in detail?
4. What are the other methods to solve cubic equations?
5. How do you factor quadratic polynomials?
6. What are the applications of finding roots of polynomials?
7. How do complex roots of polynomials affect their graphs?
8. Can a cubic equation have more than three real roots?

### Tip
Always check the possible rational roots first to simplify the factorization process of a polynomial equation.

Learn how to solve the cubic equation x^3 + 3x^2 - 4x - 12 with detailed step-by-step instructions. This problem covers polynomial equations, the Rational Root Theorem, synthetic division, and factoring techniques. Suitable for students in Grades 10-12.

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