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 = -