Let's find all the zeros of the function $f(x) = x^3 - 4x^2 + 9x - 36$ using synthetic division.

Step 1: Trial and Error for Possible Rational Zeros

The first step is to check for possible rational zeros using the Rational Root Theorem. According to the Rational Root Theorem, the potential rational zeros are the factors of the constant term ($-36$) divided by the factors of the leading coefficient ($1$).

Factors of $-36$: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

Factors of $1$ (the leading coefficient): $\pm 1$

So the possible rational zeros are: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

Step 2: Testing Rational Zeros

We will now test some of these potential zeros using synthetic division. Let's start by testing $x = 3$.

Step 3: Perform Synthetic Division (test $x = 3$)

We will divide the polynomial $x^3 - 4x^2 + 9x - 36$ by $x - 3$ using synthetic division.

The coefficients of $f(x)$ are: $1, -4, 9, -36$

Set up synthetic division with $x = 3$:

3 & 1 & -4 & 9 & -36 \\ & & 3 & -3 & 18 \\ \hline & 1 & -1 & 6 & 0 \\ \end{array}$$ The remainder is $$0$$, which means $$x = 3$$ is indeed a root. ### Step 4: Factor the Polynomial After synthetic division, the quotient is: $$x^2 - x + 6$$ Thus, the factorization of $$f(x)$$ is: $$f(x) = (x - 3)(x^2 - x + 6)$$ ### Step 5: Solve for Remaining Zeros Now, we solve for the zeros of $$x^2 - x + 6$$ by using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For $$x^2 - x + 6$$, $$a = 1$$, $$b = -1$$, and $$c = 6$$. $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(6)}}{2(1)} = \frac{1 \pm \sqrt{1 - 24}}{2} = \frac{1 \pm \sqrt{-23}}{2}$$ $$x = \frac{1 \pm i\sqrt{23}}{2}$$ So the remaining roots are complex: $$x = \frac{1 + i\sqrt{23}}{2}, \quad x = \frac{1 - i\sqrt{23}}{2}$$ ### Final Answer: The zeros of the polynomial $$f(x) = x^3 - 4x^2 + 9x - 36$$ are: $$x = 3, \quad x = \frac{1 + i\sqrt{23}}{2}, \quad x = \frac{1 - i\sqrt{23}}{2}$$ ### Additional Notes: - $$x = 3$$ is a real root. - The other two roots are complex conjugates. Would you like any further details on synthetic division or the quadratic formula? Here are some additional questions related to this topic: 1. How do you use synthetic division to check for other possible rational roots? 2. Can synthetic division be used for higher-degree polynomials (degree 4 or more)? 3. What happens when a polynomial has a complex root? How does it affect the division process? 4. How does the Rational Root Theorem help in narrowing down the number of trials needed? 5. What is the connection between the discriminant of a quadratic and the nature of its roots? **Tip:** Always start with the simplest possible rational root (like $$x = 1$$ or $$x = -1$$) to quickly eliminate obvious candidates before diving into more complicated calculations.