how to solve this equation in the real number system:

2x^(3) + 3x^(2) + 2x + 3 = 0

steps please

2X^3 + 3X^2 + 2X + 3 = 0.

First, we reduce the cubic Eq to a guadratic Eq by finding a value of x that satisfies the Eq. This was done by
trial and error. However, the task was made a lot easier by the use of EXCEL Spread sheets.

It was determined x = -1.5 satisfies the cubic Eq:

x = -1.5 = -1 1/2 = -3/2.

x = -3/2,
x + 3/2 = 0,
Multiply both sides by 2 and get:
2x + 3 = 0.
Using long division, we divide the cubic Eq by 2x + 3:

(2x^3 + 3x^2 +2x + 3)/(2x + 3)=
x^2 + 1.
Now we can easily solve the quad. Eq:

x^2 + 1 = 0,

x2 = -1,
x = sqrt(-1) = +- i.

Solution set: x = - 3/2, x = i, x = -i.

So there are 3 solutions which is the max. for a cubic Eq.

Well, this equation looks like quite a party! Let's see if we can find some real solutions to bring everyone back to reality.

Step 1: Start by factoring out any common factors. In this equation, we have no common factors to factor out. So, we move on.

Step 2: Look for any rational roots using the rational root theorem. The rational root theorem suggests that any rational root can be written as a fraction where the numerator divides evenly into the constant term (in this case, 3) and the denominator divides evenly into the leading coefficient (in this case, 2). After checking all possible combinations, we find that there are no rational roots.

Well, looks like this equation doesn't want to give us any real solutions. It's quite the prankster! Sometimes equations can be a bit mischievous like that. Maybe try some other methods or techniques to see if we can get closer to a solution.

To solve the equation 2x^(3) + 3x^(2) + 2x + 3 = 0 in the real number system, follow these steps:

Step 1: Start by applying the rational root theorem to find possible rational roots. The rational root theorem states that if a rational number "p/q" is a root of the equation, then p must be a factor of the constant term (in this case, 3), and q must be a factor of the leading coefficient (in this case, 2).

The factors of 3 are ±1 and ±3, and the factors of 2 are ±1 and ±2. Therefore, the possible rational roots are ±1, ±3, ±1/2, and ±3/2.

Step 2: Test each of the possible rational roots using synthetic division or substitution. Start with any of the possible rational roots and check if it satisfies the equation by substituting it into the equation.

For example, let's try x = 1 as a possible root:

2(1)^(3) + 3(1)^(2) + 2(1) + 3 = 0

2 + 3 + 2 + 3 = 0

Since this does not equal zero, x = 1 is not a root.

Step 3: Repeat Step 2 for each of the possible rational roots until you find one that satisfies the equation.

Continuing with the example, let's try x = -1 as a possible root:

2(-1)^(3) + 3(-1)^(2) + 2(-1) + 3 = 0

-2 + 3 - 2 + 3 = 0

Since this does equal zero, x = -1 is a root.

Step 4: Once you find a root, divide the equation by (x - root) using synthetic division or long division. In this case, (x + 1) is a factor because x = -1 is the root we found.

Performing synthetic division with the equation divided by (x + 1):

2 | 2 3 2 3
-2 -2 0
___________
0 1 2 3

The result is a remainder of zero, and the quotient is 2x^(2) + x + 3.

Step 5: The resulting quotient, 2x^(2) + x + 3, is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula.

In this case, the quadratic equation 2x^(2) + x + 3 = 0 does not factor easily, so we will use the quadratic formula:

x = (-b ± √(b^(2) - 4ac)) / 2a

For our equation, a = 2, b = 1, and c = 3. Substituting these values into the quadratic formula:

x = (-1 ± √(1 - 4(2)(3))) / (2(2))

Simplifying further:

x = (-1 ± √(1 - 24)) / 4

x = (-1 ± √(-23)) / 4

Since the square root of a negative number does not exist in the real number system, there are no real solutions for the quadratic equation.

Therefore, the solution to the original equation 2x^(3) + 3x^(2) + 2x + 3 = 0 in the real number system is x = -1.

To solve the given equation, we can use a method called the Rational Root Theorem to find the possible rational roots of the equation. Once we find these potential roots, we can test each one to see if it satisfies the equation and find the real solutions.

Step 1: Apply the Rational Root Theorem
The Rational Root Theorem states that any rational root of the equation has the form p/q, where p is a factor of the constant term (in this case, 3) and q is a factor of the leading coefficient (in this case, 2). So, the possible rational roots of our equation are ±1, ±3, ±1/2, ±3/2.

Step 2: Test the potential rational roots
For each potential rational root, substitute the value into the equation and check if it equals zero. Starting with the easiest values, begin substituting until you find a root. Substitute one root at a time and check if it solves the equation.

Substituting x = 1: (2(1)^3 + 3(1)^2 + 2(1) + 3) = 2 + 3 + 2 + 3 = 10 ≠ 0
Substituting x = -1: (2(-1)^3 + 3(-1)^2 + 2(-1) + 3) = -2 + 3 - 2 + 3 = 2 ≠ 0
Substituting x = 3: (2(3)^3 + 3(3)^2 + 2(3) + 3) = 54 + 27 + 6 + 3 = 90 ≠ 0
Substituting x = -3: (2(-3)^3 + 3(-3)^2 + 2(-3) + 3) = -54 + 27 - 6 + 3 = -30 ≠ 0
Substituting x = 1/2: (2(1/2)^3 + 3(1/2)^2 + 2(1/2) + 3) = 1/4 + 3/4 + 1 + 3 = 17/4 ≠ 0
Substituting x = -1/2: (2(-1/2)^3 + 3(-1/2)^2 + 2(-1/2) + 3) = -1/4 + 3/4 - 1 + 3 = 6/4 ≠ 0
Substituting x = 3/2: (2(3/2)^3 + 3(3/2)^2 + 2(3/2) + 3) = 27/4 + 27/4 + 3 + 3 = 63/4 ≠ 0
Substituting x = -3/2: (2(-3/2)^3 + 3(-3/2)^2 + 2(-3/2) + 3) = -27/4 + 27/4 - 3 + 3 = 0

Step 3: Find the real roots
From the potential rational roots that we tested, we found that x = -3/2 satisfies the equation. Therefore, it is the real root of the equation 2x^(3) + 3x^(2) + 2x + 3 = 0.

So, the real solution to the equation is x = -3/2.