Finding Exact Zeros Of F(x) = X^3 - 3x^2 - 3x + 1 A Step-by-Step Guide

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Finding the zeros of a polynomial function is a fundamental problem in algebra. The zeros, also known as roots, are the values of $x$ for which the function $f(x)$ equals zero. In this comprehensive guide, we will delve into the process of finding the zeros of the cubic function $f(x) = x^3 - 3x^2 - 3x + 1$. This function, while appearing complex at first glance, can be solved using a combination of algebraic techniques and trigonometric identities. Our focus will be on obtaining exact values, avoiding decimal approximations, to ensure precision and a deeper understanding of the solution. Understanding how to solve cubic equations is essential in various fields, including engineering, physics, and computer science, where polynomial functions are used to model real-world phenomena. This article will provide a step-by-step approach, making it accessible for students, educators, and anyone interested in enhancing their mathematical problem-solving skills. The journey to finding these zeros will not only showcase the beauty of mathematical methods but also highlight the interconnectedness of different mathematical concepts. We will start by exploring possible rational roots and then transition to trigonometric methods to extract the precise solutions. This blend of algebraic and trigonometric techniques underscores the versatility required in advanced mathematical problem-solving.

1. Initial Assessment and Rational Root Theorem

When faced with a cubic function like $f(x) = x^3 - 3x^2 - 3x + 1$, the first step is to assess its characteristics. We observe that it is a cubic polynomial, meaning it has a degree of 3, and thus, it can have up to three roots. The Rational Root Theorem is a useful tool to begin our search for rational roots. This theorem states that if a polynomial has integer coefficients, then any rational root must be of the form $\pm \frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. In our case, the constant term is 1 and the leading coefficient is also 1. Therefore, the possible rational roots are $\pm 1$. Let's test these values by substituting them into the function:

• $f(1) = (1)^3 - 3(1)^2 - 3(1) + 1 = 1 - 3 - 3 + 1 = -4 \neq 0$
• $f(-1) = (-1)^3 - 3(-1)^2 - 3(-1) + 1 = -1 - 3 + 3 + 1 = 0$

We find that $f(-1) = 0$, which means that $x = -1$ is a root of the equation. This discovery is crucial because it allows us to factor the cubic polynomial, reducing it to a quadratic equation, which is easier to solve. Factoring the polynomial is a key step in simplifying the problem and making it more manageable. By identifying one rational root, we pave the way for finding the remaining roots through further algebraic manipulation. The ability to apply the Rational Root Theorem efficiently can significantly streamline the process of solving polynomial equations. This initial assessment not only helps in identifying potential roots but also provides a foundational understanding of the polynomial's behavior.

2. Polynomial Division and Quadratic Reduction

Since we've identified $x = -1$ as a root of $f(x) = x^3 - 3x^2 - 3x + 1$, we know that $(x + 1)$ is a factor of the polynomial. To find the other factor, we perform polynomial division. Dividing $x^3 - 3x^2 - 3x + 1$ by $(x + 1)$ yields a quadratic polynomial. Polynomial long division is a systematic way to divide polynomials, similar to long division with numbers. It allows us to break down the cubic polynomial into a linear factor and a quadratic factor. This step is crucial because quadratic equations have well-established methods for finding their roots. The result of the division is:

        x^2 - 4x + 1
x + 1 | x^3 - 3x^2 - 3x + 1
        - (x^3 + x^2)
        -----------
              -4x^2 - 3x
              - (-4x^2 - 4x)
              -------------
                      x + 1
                      - (x + 1)
                      -------
                          0

Therefore, $f(x)$ can be written as $(x + 1)(x^2 - 4x + 1)$. Now, we have reduced the problem to finding the roots of the quadratic equation $x^2 - 4x + 1 = 0$. This quadratic equation can be solved using the quadratic formula or by completing the square. Reducing the cubic polynomial to a quadratic equation is a significant simplification, as quadratic equations are much easier to handle. This step demonstrates the power of factoring and polynomial division in solving higher-degree polynomial equations. The resulting quadratic equation holds the key to finding the remaining two roots of the original cubic function. By transitioning from a cubic to a quadratic, we leverage our knowledge of quadratic equation-solving techniques to progress towards the final solution.

3. Solving the Quadratic Equation

Now, we focus on solving the quadratic equation $x^2 - 4x + 1 = 0$. We can use the quadratic formula, which is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$. In our case, $a = 1$, $b = -4$, and $c = 1$. Substituting these values into the quadratic formula, we get:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)}$$

$$x = \frac{4 \pm \sqrt{16 - 4}}{2}$$

$$x = \frac{4 \pm \sqrt{12}}{2}$$

$$x = \frac{4 \pm 2\sqrt{3}}{2}$$

$$x = 2 \pm \sqrt{3}$$

Thus, the two roots of the quadratic equation are $x = 2 + \sqrt{3}$ and $x = 2 - \sqrt{3}$. These roots, along with the root $x = -1$ that we found earlier, constitute the complete set of roots for the cubic function. The quadratic formula is a fundamental tool in algebra, providing a direct method for solving quadratic equations. Its application here demonstrates its versatility and importance in solving polynomial equations. The roots obtained, $2 + \sqrt{3}$ and $2 - \sqrt{3}$, are irrational numbers, highlighting that not all polynomial roots are rational. This step underscores the importance of having a robust method like the quadratic formula to find roots that may not be easily discernible. The exact values of these roots provide a precise solution to the equation, avoiding the inaccuracies that can arise from decimal approximations.

4. Summarizing the Zeros

We have successfully found all the zeros of the cubic function $f(x) = x^3 - 3x^2 - 3x + 1$. These zeros are the values of $x$ for which $f(x) = 0$. We found one rational root using the Rational Root Theorem and then used polynomial division to reduce the cubic to a quadratic. Finally, we applied the quadratic formula to find the remaining two roots. The zeros are:

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

These are the exact values of the zeros, expressed without decimal approximations. Listing the zeros explicitly provides a clear and concise answer to the problem. The combination of algebraic techniques used – the Rational Root Theorem, polynomial division, and the quadratic formula – showcases a comprehensive approach to solving polynomial equations. Each root contributes to the complete solution of the problem, representing a value where the function intersects the x-axis. The ability to find exact values is crucial in many mathematical and scientific contexts, where precision is paramount. This summary not only presents the solutions but also reinforces the methodical approach used to obtain them.

5. Alternative Trigonometric Approach (Advanced)

While we have found the zeros using algebraic methods, it's worth noting that there's also a trigonometric approach to solving cubic equations, which can be particularly elegant for certain forms. This method involves using trigonometric identities to transform the cubic equation into a form that can be solved using trigonometric functions. Let's consider the substitution $x = y + 1$ in the original equation $f(x) = x^3 - 3x^2 - 3x + 1$:

$$f(y + 1) = (y + 1)^3 - 3(y + 1)^2 - 3(y + 1) + 1$$

$$= y^3 + 3y^2 + 3y + 1 - 3(y^2 + 2y + 1) - 3y - 3 + 1$$

$$= y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 - 3y - 3 + 1$$

$$= y^3 - 6y - 4$$

Now we have a depressed cubic equation of the form $y^3 + py + q = 0$, where $p = -6$ and $q = -4$. To solve this trigonometrically, we make the substitution $y = 2\sqrt{-p/3} \cos(\theta)$, where $-p/3 = -(-6)/3 = 2$. Thus, $y = 2\sqrt{2} \cos(\theta)$. Substituting this into the depressed cubic equation, we get:

$$(2\sqrt{2} \cos(\theta))^3 - 6(2\sqrt{2} \cos(\theta)) - 4 = 0$$

$$16\sqrt{2} \cos^3(\theta) - 12\sqrt{2} \cos(\theta) - 4 = 0$$

Divide by $4\sqrt{2}$:

$$4\cos^3(\theta) - 3\cos(\theta) - \sqrt{2} = 0$$

Using the triple angle identity $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$, we have:

$$\cos(3\theta) = \frac{\sqrt{2}}{2}$$

$$3\theta = \pm \frac{\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}$$

$$\theta = \pm \frac{\pi}{12} + \frac{2\pi k}{3}, \quad k = 0, 1, 2$$

For $k = 0$:

$$\theta_1 = \frac{\pi}{12}$$

$$y_1 = 2\sqrt{2} \cos(\frac{\pi}{12}) = 2\sqrt{2} \frac{\sqrt{6} + \sqrt{2}}{4} = 1 + \sqrt{3}$$

$$x_1 = y_1 + 1 = 2 + \sqrt{3}$$

For $k = 1$:

$$\theta_2 = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{9\pi}{12} = \frac{3\pi}{4}$$

$$y_2 = 2\sqrt{2} \cos(\frac{3\pi}{4}) = 2\sqrt{2} (-\frac{\sqrt{2}}{2}) = -2$$

$$x_2 = y_2 + 1 = -1$$

For $k = 2$:

$$\theta_3 = \frac{\pi}{12} + \frac{4\pi}{3} = \frac{17\pi}{12}$$

$$y_3 = 2\sqrt{2} \cos(\frac{17\pi}{12}) = 2\sqrt{2} \cos(\frac{17\pi}{12}) = -1 - \sqrt{3}$$

$$x_3 = y_3 + 1 = -\sqrt{3}$$

This trigonometric approach provides an alternative method to find the roots and further illustrates the rich interconnectedness of mathematical concepts. While more complex, it highlights the versatility of trigonometric identities in solving algebraic problems.

In conclusion, we have explored both algebraic and trigonometric methods to find the zeros of the cubic function $f(x) = x^3 - 3x^2 - 3x + 1$, demonstrating a comprehensive approach to solving polynomial equations.

Zeros: -1, $2 + \sqrt{3}$, $2 - \sqrt{3}$