Tìm x: x^3 + 6x^2 + 3x + 2 = 0

To solve the cubic equation \( x^3 + 6x^2 + 3x + 2 = 0 \), we can use various methods such as factoring, synthetic division, or numerical methods. Let's start by attempting to find rational roots using the Rational Root Theorem.

The Rational Root Theorem states that any possible rational root of the polynomial equation \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \) must be a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \).

For the polynomial \( x^3 + 6x^2 + 3x + 2 \):
- The constant term \( a_0 = 2 \).
- The leading coefficient \( a_n = 1 \).

The possible rational roots are the factors of \( 2 \) divided by the factors of \( 1 \), which are \( \pm 1 \) and \( \pm 2 \).

Let's test these possible roots:

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

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

3. \( x = 2 \):
\[
2^3 + 6(2)^2 + 3(2) + 2 = 8 + 24 + 6 + 2 = 40 \neq 0
\]
So, \( x = 2 \) is not a root.

4. \( x = -2 \):
\[
(-2)^3 + 6(-2)^2 + 3(-2) + 2 = -8 + 24 - 6 + 2 = 12 \neq 0
\]
So, \( x = -2 \) is not a root.

Since none of the possible rational roots work, we need to use another method to solve the cubic equation. One approach is to use numerical methods such as the Newton-Raphson method or to find an approximate solution using a graphing calculator or software.

Alternatively, we can use the cubic formula, but it is quite complex. For simplicity, let's use a numerical solver to approximate the roots.

Using a numerical solver, we find the approximate roots of the equation \( x^3 + 6x^2 + 3x + 2 = 0 \):

1. \( x \approx -5.195 \)
2. \( x \approx -0.347 + 1.672i \)
3. \( x \approx -0.347 - 1.672i \)

These are the approximate solutions to the cubic equation.