Answer
The real zeros of the polynomial are $x = -1$ or $x = \frac{\sqrt{13}}{3} - \frac{1}{3}$ or $x = -\frac{1}{3} - \frac{\sqrt{13}}{3}$
Work Step by Step
Solve for x over the real numbers:
$3 x^3 + 5 x^2 - 2 x - 4 = 0$
The left hand side factors into a product with two terms:
$(x + 1) (3 x^2 + 2 x - 4) = 0$
Split into two equations:
$x + 1 = 0$ or $3 x^2 + 2 x - 4 = 0$
Subtract 1 from both sides:
$x = -1$ or $3 x^2 + 2 x - 4 = 0$
Divide both sides by 3:
$x = -1$ or $x^2 + \frac{(2 x)}{3} - \frac{4}{3} = 0$
Add $\frac{4}{3}$ to both sides:
$x = -1$ or $x^2 + \frac{(2 x)}{3} = \frac{4}{3}$
Add $\frac{1}{9}$ to both sides:
$x = -1$ or $x^2 + \frac{(2 x)}{3} + \frac{1}{9} = \frac{13}{9}$
Write the left hand side as a square:
$x = -1$ or $(x + \frac{1}{3})^2 = \frac{13}{9}$
Take the square root of both sides:
$x = -1$ or $x + \frac{1}{3} = \frac{\sqrt{13}}{3}$ or $x + \frac{1}{3} = -\frac{\sqrt{13}}{3}$
Subtract $\frac{1}{3}$ from both sides:
$x = -1$ or $x = \frac{\sqrt{13}}{3} - \frac{1}{3}$ or $x + \frac{1}{3} = -\frac{\sqrt{13}}{3}$
Answer:
The real zeros of the polynomial are $x = -1$ or $x = \frac{\sqrt{13}}{3} - \frac{1}{3}$ or $x = -\frac{1}{3} - \frac{\sqrt{13}}{3}$
