To find the real zeros of the function $f(x) = 2x^3 - x^2 - 8x + 4$, we need to solve the equation:

$f(x) = 2x^3 - x^2 - 8x + 4 = 0$

We can use the Rational Root Theorem to test potential rational solutions based on the factors of the constant term (4) and the leading coefficient (2). The possible rational roots are $\pm 1, \pm 2, \pm \frac{1}{2}, \pm 4$.

Step 1: Test $x = 1$

Substitute $x = 1$ into the equation:

$f(1) = 2(1)^3 - (1)^2 - 8(1) + 4 = 2 - 1 - 8 + 4 = -3 \quad (\text{not zero})$

Step 2: Test $x = -1$

Substitute $x = -1$ into the equation:

$f(-1) = 2(-1)^3 - (-1)^2 - 8(-1) + 4 = -2 - 1 + 8 + 4 = 9 \quad (\text{not zero})$

Step 3: Test $x = 2$

Substitute $x = 2$ into the equation:

$f(2) = 2(2)^3 - (2)^2 - 8(2) + 4 = 16 - 4 - 16 + 4 = 0 \quad (\text{zero found!})$

Step 4: Test $x = -2$

Substitute $x = -2$ into the equation:

$f(-2) = 2(-2)^3 - (-2)^2 - 8(-2) + 4 = -16 - 4 + 16 + 4 = 0 \quad (\text{zero found!})$

Step 5: Test $x = \frac{1}{2}$

Substitute $x = \frac{1}{2}$ into the equation:

$f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - 8\left(\frac{1}{2}\right) + 4 = 2\left(\frac{1}{8}\right) - \frac{1}{4} - 4 + 4 = \frac{1}{4} - \frac{1}{4} - 4 + 4 = 0 \quad (\text{zero found!})$

Conclusion

The real zeros of the function are $x = 2, x = -2, \text{ and } x = \frac{1}{2}$. Therefore, the correct answers are:

• $x = 2$
• $x = -2$
• $x = \frac{1}{2}$

Would you like more details on this solution?

Here are five related questions:

1. How can the Rational Root Theorem be applied to higher-degree polynomials?
2. What are the different methods to find the roots of a cubic function?
3. How does synthetic division help in finding zeros of polynomials?
4. What is the role of the discriminant in identifying the nature of polynomial roots?
5. Can graphing help verify the solutions to polynomial equations?

Tip: Always check for potential rational roots using simple substitutions to quickly narrow down your search for zeros.