The function given in the image is:

$f(x) = x^3 - 5x^2 + 5x + 3$

To find the zeros of this cubic function, we need to solve $f(x) = 0$, i.e.,

$x^3 - 5x^2 + 5x + 3 = 0$

Steps:

1. Try factoring: Since this is a cubic equation, we can start by using the Rational Root Theorem to test possible rational roots (factors of the constant term over the leading coefficient).

   • The constant term is $3$, and the leading coefficient is $1$. Possible rational roots are $\pm 1, \pm 3$.

2. Test $x = 1$ by substituting it into the equation: $f(1) = 1^3 - 5(1)^2 + 5(1) + 3 = 1 - 5 + 5 + 3 = 4$ Since $f(1) \neq 0$, $x = 1$ is not a root.

3. Test $x = -1$: $f(-1) = (-1)^3 - 5(-1)^2 + 5(-1) + 3 = -1 - 5 - 5 + 3 = -8$ So, $f(-1) \neq 0$.

4. Test $x = 3$: $f(3) = 3^3 - 5(3)^2 + 5(3) + 3 = 27 - 45 + 15 + 3 = 0$ So, $x = 3$ is a root.

5. Use synthetic division: Now that we know $x = 3$ is a root, divide $f(x)$ by $(x - 3)$ using synthetic division to factor the cubic polynomial.

After synthetic division, the quotient is $x^2 - 2x - 1$.

6. Solve the quadratic: $x^2 - 2x - 1 = 0$ Using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

Final Zeros:

Thus, the zeros of $f(x)$ are:

$x = 3, 1 + \sqrt{2}, 1 - \sqrt{2}$

Would you like further explanation or details on any step?

Here are 5 questions related to this topic:

1. What is the Rational Root Theorem and how is it applied?
2. How do you perform synthetic division on polynomials?
3. Can all cubic equations be factored or solved this way?
4. What are some other methods to find roots of polynomials?
5. How can you verify that the roots of a polynomial are correct?

Tip: When solving polynomials, always check simple rational roots first, as they can help simplify more complex equations.