Question

$3t^3 + 4t^2 - 5t - 2 = 0$

Answer 1

$t = 1, \quad t = -2, \quad t = -\frac{1}{3}$

Answer 2

Given the cubic equation:

$$3t^3 + 4t^2 - 5t - 2 = 0$$

1. Find a rational root:
   By the Rational Root Theorem, test possible roots ±1, ±2, ±1/3, ±2/3.
   Testing $t = 1$:

   $$3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0.$$

   Hence, $t = 1$ is a root.

2. Factor the polynomial:
   Factor out $(t - 1)$ from the polynomial. Dividing by $t - 1$ (using synthetic or polynomial division), we get:

   $$3t^3 + 4t^2 - 5t - 2 = (t-1)(3t^2 + 7t + 2).$$

3. Solve the quadratic:
   Factor the quadratic:

   $$3t^2 + 7t + 2 = 0.$$

   Notice that:

   $$3t^2 + 6t + t + 2 = 3t(t+2) + 1(t+2) = (t+2)(3t+1)=0.$$

   Set each factor to zero:

   $$t + 2 = 0 \quad \Rightarrow \quad t = -2,$$ $$3t + 1 = 0 \quad \Rightarrow \quad t = -\frac{1}{3}.$$

4. Final Answer:
   The solutions are:

   $$t = 1, \quad t = -2, \quad t = -\frac{1}{3}.$$

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Minimal Explanation:
Find $t=1$ by testing, factor to get $(t-1)(3t^2+7t+2)=0$, then factor the quadratic as $(t+2)(3t+1)=0$ to obtain the remaining roots $t=-2$ and $t=-\frac{1}{3}$.