You provided the equation:

\[
2x^3 + 3x^2 - 1 = 0
\]

This is a cubic equation. Let's explore possible methods for solving it.

### Step 1: Try Rational Root Theorem

The Rational Root Theorem helps identify potential rational solutions of the form \( \frac{p}{q} \), where:
- \( p \) is a factor of the constant term (here, \(-1\)),
- \( q \) is a factor of the leading coefficient (here, \(2\)).

Possible values for \( x \) are:

\[
x = \pm 1, \pm \frac{1}{2}
\]

### Step 2: Test possible values of \( x \)

We will substitute these values into the equation to check for valid solutions.

#### Test \( x = 1 \):

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

#### Test \( x = -1 \):

\[
2(-1)^3 + 3(-1)^2 - 1 = 2(-1) + 3(1) - 1 = -2 + 3 - 1 = 0
\]
So, \( x = -1 \) **is a solution**.

### Step 3: Factor the cubic equation

Since \( x = -1 \) is a solution, we can now factor the cubic equation using \( (x + 1) \) as one of the factors.

Using synthetic division or polynomial long division, divide \( 2x^3 + 3x^2 - 1 \) by \( (x + 1) \):

\[
\frac{2x^3 + 3x^2 - 1}{x + 1} = 2x^2 + x - 1
\]

So, the factored form of the equation is:

\[
(2x^2 + x - 1)(x + 1) = 0
\]

### Step 4: Solve the quadratic equation

Now solve the quadratic \( 2x^2 + x - 1 = 0 \) using the quadratic formula:

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

For \( 2x^2 + x - 1 = 0 \), \( a = 2 \), \( b = 1 \), and \( c = -1 \). Substituting these values into the quadratic formula:

\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}
\]
\[
x = \frac{-1 \pm \sqrt{1 + 8}}{4}
\]
\[
x = \frac{-1 \pm \sqrt{9}}{4}
\]
\[
x = \frac{-1 \pm 3}{4}
\]

So, the two solutions are:

\[
x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1
\]

### Step 5: Conclusion

The full set of solutions for the cubic equation \( 2x^3 + 3x^2 - 1 = 0 \) is:

\[
x = -

Explore the step-by-step solution to the cubic equation 2x^3 + 3x^2 - 1 = 0. This guide covers key algebraic concepts such as the Rational Root Theorem and the quadratic formula, providing a comprehensive approach to solving cubic equations. Ideal for students in Grades 9-12.

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