To find all zeros of the function $f(x) = x^3 - 4x^2 + x + 6$, we solve the equation $f(x) = 0$.

Step 1: Apply the Rational Root Theorem

The Rational Root Theorem suggests that any potential rational root is of the form: $\pm \frac{p}{q}$ where:

• $p$ is a factor of the constant term ($6$),
• $q$ is a factor of the leading coefficient ($1$).

Here, $p = \pm 1, \pm 2, \pm 3, \pm 6$, and $q = \pm 1$. Therefore, the possible rational roots are: $\pm 1, \pm 2, \pm 3, \pm 6$

Step 2: Test Potential Roots

Using synthetic or direct substitution, we test the potential roots.

Test $x = 1$:

Substitute $x = 1$ into $f(x)$: $f(1) = 1^3 - 4(1^2) + 1 + 6 = 1 - 4 + 1 + 6 = 4 \quad \text{(Not a root)}.$

Test $x = 2$:

Substitute $x = 2$: $f(2) = 2^3 - 4(2^2) + 2 + 6 = 8 - 16 + 2 + 6 = 0 \quad \text{(Root found)}.$

Step 3: Factor the Polynomial

Since $x = 2$ is a root, $x - 2$ is a factor of $f(x)$. Perform synthetic or polynomial division to factorize $f(x)$.

Synthetic Division:

2 & 1 & -4 & 1 & 6 \\ & & 2 & -4 & -6 \\ \hline & 1 & -2 & -3 & 0 \\ \end{array}$$ The quotient is $$x^2 - 2x - 3$$, so: $$f(x) = (x - 2)(x^2 - 2x - 3)$$ ### Step 4: Factorize the Quadratic Factorize $$x^2 - 2x - 3$$: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ ### Step 5: Write the Complete Factorization The complete factorization of $$f(x)$$ is: $$f(x) = (x - 2)(x - 3)(x + 1)$$ ### Step 6: Find All Zeros The zeros of $$f(x)$$ are: $$x = 2, \, x = 3, \, x = -1$$ ### Final Answer: The zeros of the function are: $$\boxed{2, 3, -1}$$ --- Would you like me to provide additional details on any of the steps? Here are 5 related questions to expand on this concept: 1. How do you verify the roots of a polynomial after factorization? 2. What is the significance of the Rational Root Theorem in solving polynomials? 3. How would you solve this polynomial if no rational roots existed? 4. Can you demonstrate synthetic division with another example? 5. How do the degree of the polynomial and the number of its roots relate? **Tip:** Always test all potential roots systematically to avoid missing any rational solutions.