To find the real zeros of the polynomial $f(x) = 3x^4 - x^3 - 6x^2$, we will follow these steps:

1. Factor the polynomial as much as possible.
2. Set each factor equal to zero to find the real zeros.

Step 1: Factor the Polynomial

First, factor out the greatest common factor (GCF), which is $x^2$:

$f(x) = x^2(3x^2 - x - 6)$

Now, we have $f(x) = x^2(3x^2 - x - 6)$.

Step 2: Set Each Factor Equal to Zero

We have two factors: $x^2$ and $3x^2 - x - 6$.

Factor 1: $x^2 = 0$

$x = 0$ This gives us a zero at $x = 0$.

Factor 2: $3x^2 - x - 6 = 0$

We will solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 3$, $b = -1$, and $c = -6$. Substituting these values, we get:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-6)}}{2(3)}$

$x = \frac{1 \pm \sqrt{1 + 72}}{6}$

$x = \frac{1 \pm \sqrt{73}}{6}$

This gives two real zeros:

$x = \frac{1 + \sqrt{73}}{6}, \quad x = \frac{1 - \sqrt{73}}{6}$

Conclusion

The real zeros of the polynomial $f(x) = 3x^4 - x^3 - 6x^2$ are:

$0, \quad \frac{1 + \sqrt{73}}{6}, \quad \frac{1 - \sqrt{73}}{6}$

If you have any questions or need more details, feel free to ask!

Here are 5 related questions for further practice:

1. How would you find the zeros of $f(x) = x^3 - 4x$?
2. Can you factor $x^4 - 16$ into real factors?
3. What are the zeros of the polynomial $g(x) = 2x^3 - 3x^2 - 11x + 6$?
4. How do you solve $4x^2 + 5x - 6 = 0$ using the quadratic formula?
5. Can you find all real zeros of $h(x) = 2x^4 - 3x^2 + 1$?

Tip: When factoring a polynomial, always start by looking for a greatest common factor (GCF).