https://file.aiquickdraw.com/mathbot/file/1727096885065dxu9ni8d.jpg

Let's solve the two problems shown in the image:

### Problem 3: \( f(x) = x^3 - 3x^2 - x + 3 \)

1. **Number of possible real zeros:**
   - The degree of the polynomial is 3, so there can be up to 3 real zeros.

2. **Number of turning points:**
   - The number of turning points is at most \( n-1 \), where \( n \) is the degree of the polynomial.
   - Thus, this cubic function can have up to \( 3 - 1 = 2 \) turning points.

3. **Factoring to find the real zeros:**
   Let's try factoring the cubic polynomial:
   \[
   f(x) = x^3 - 3x^2 - x + 3
   \]
   We'll use the Rational Root Theorem to test possible rational roots (which are factors of 3):
   Possible rational roots: \( \pm 1, \pm 3 \)

   Testing \( x = 1 \):
   \[
   f(1) = 1^3 - 3(1^2) - 1 + 3 = 1 - 3 - 1 + 3 = 0
   \]
   So, \( x = 1 \) is a root.

   Now divide \( f(x) \) by \( (x - 1) \) using synthetic division:
   \[
   \frac{x^3 - 3x^2 - x + 3}{x - 1} = x^2 - 2x - 3
   \]

   Factor \( x^2 - 2x - 3 \):
   \[
   x^2 - 2x - 3 = (x - 3)(x + 1)
   \]

   Therefore, the fully factored form of the function is:
   \[
   f(x) = (x - 1)(x - 3)(x + 1)
   \]

   **Real Zeros:**
   - \( x = 1 \)
   - \( x = 3 \)
   - \( x = -1 \)

### Problem 4: \( f(x) = x^4 - 5x^2 - 36 \)

1. **Number of possible real zeros:**
   - The degree of the polynomial is 4, so there can be up to 4 real zeros.

2. **Number of turning points:**
   - The number of turning points is at most \( 4 - 1 = 3 \).

3. **Factoring to find the real zeros:**
   Let’s use substitution to factor:
   Set \( u = x^2 \), then:
   \[
   f(x) = u^2 - 5u - 36
   \]

   Factor the quadratic:
   \[
   u^2 - 5u - 36 = (u - 9)(u + 4)
   \]

   Substituting \( u = x^2 \) back:
   \[
   (x^2 - 9)(x^2 + 4) = 0
   \]

   Factor \( x^2 - 9 \):
   \[
   x^2 - 9 = (x - 3)(x + 3)
   \]

   So the factored form is:
   \[
   (x - 3)(x + 3)(x^2 + 4) = 0
   \]

   **Real Zeros:**
   - \( x = 3 \)
   - \( x = -3 \)

   The factor \( x^2 + 4 \) does not yield real roots because it gives imaginary solutions \( x = \pm 2i \).

   **Summary:**
   - The real zeros are \( x = 3 \) and \( x = -3 \).

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Would you like more details or have any questions?

### Related Questions:
1. How can I graph the polynomial functions to visualize the turning points?
2. What does the Rational Root Theorem tell us in polynomial factorization?
3. How do imaginary solutions impact the overall graph of the polynomial?
4. Why can a degree-3 polynomial have at most 2 turning points?
5. How does synthetic division help simplify higher-degree polynomials?

**Tip:** Always test potential rational roots from the Rational Root Theorem before factoring a cubic or higher-degree polynomial.

State the number of possible real zeros and turning points of each function. Then determine all of the real zeros by factoring.

f(x) = x^3 - 3x^2 - x + 3
f(x) = x^4 - 5x^2 - 36

Learn how to factor cubic and quartic polynomials such as x^3 - 3x^2 - x + 3 and x^4 - 5x^2 - 36. Discover the real zeros, turning points, and apply the Rational Root Theorem and synthetic division for solving these polynomials. Suitable for Grades 9-12.

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