To find the real zeros of the function $f(x) = x^3 - 7x^2 - 41x - 33$, we need to identify values of $x$ for which $f(x) = 0$.

Step 1: Use the Rational Root Theorem

The Rational Root Theorem suggests that if there are any rational roots, they will be factors of the constant term (-33) divided by factors of the leading coefficient (1). Therefore, possible rational roots are: $\pm 1, \pm 3, \pm 11, \pm 33$

Step 2: Test Possible Roots Using Synthetic Division

We can use synthetic division to test these possible values to see if any yield a remainder of zero, indicating a root.

1. Testing $x = 3$:

   • Synthetic division shows $f(3) = 0$, so $x = 3$ is a root.

2. After confirming $x = 3$ is a root, we can factor $f(x)$ as: $f(x) = (x - 3)(x^2 - 4x - 11)$

Step 3: Factor the Quadratic Expression

Now, we solve for the roots of $x^2 - 4x - 11 = 0$ using the quadratic formula: $x = \frac{4 \pm \sqrt{16 + 44}}{2} = \frac{4 \pm \sqrt{60}}{2} = \frac{4 \pm 2\sqrt{15}}{2} = 2 \pm \sqrt{15}$

Final Answer

The real zeros of $f(x)$ are: $x = 3, \quad x = 2 + \sqrt{15}, \quad x = 2 - \sqrt{15}$

Would you like further details on any of these steps, or have any other questions?

Further Questions

1. How is synthetic division performed on polynomials?
2. Why is the Rational Root Theorem useful for finding roots of polynomials?
3. What other methods can we use if the Rational Root Theorem doesn’t provide a solution?
4. How does factoring help in solving polynomial equations?
5. Can complex roots exist for this polynomial, and how would we find them?

Tip: When stuck, always try the Rational Root Theorem first for polynomials with integer coefficients. It often provides a straightforward way to find potential roots.