Factor $f(x) = 3x^3 - 2x^2 - 7x - 2$: A Step-by-Step Guide

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Hey guys! Let's dive into factorizing the cubic polynomial f(x)=3x3βˆ’2x2βˆ’7xβˆ’2f(x) = 3x^3 - 2x^2 - 7x - 2. We know that (3x+1)(3x + 1) is a factor, and our mission is to factorize f(x)f(x) fully. Buckle up; it's gonna be a fun ride!

Using the Factor Theorem

Since we know that (3x+1)(3x + 1) is a factor of f(x)f(x), we can use this information to find one of the roots of the polynomial. Setting 3x+1=03x + 1 = 0, we find that x=βˆ’13x = -\frac{1}{3} is a root. This means that f(βˆ’13)=0f(-\frac{1}{3}) = 0. Let's verify this to be sure:

f(βˆ’13)=3(βˆ’13)3βˆ’2(βˆ’13)2βˆ’7(βˆ’13)βˆ’2f(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 2(-\frac{1}{3})^2 - 7(-\frac{1}{3}) - 2

f(βˆ’13)=3(βˆ’127)βˆ’2(19)+73βˆ’2f(-\frac{1}{3}) = 3(-\frac{1}{27}) - 2(\frac{1}{9}) + \frac{7}{3} - 2

f(βˆ’13)=βˆ’19βˆ’29+219βˆ’189f(-\frac{1}{3}) = -\frac{1}{9} - \frac{2}{9} + \frac{21}{9} - \frac{18}{9}

f(βˆ’13)=βˆ’1βˆ’2+21βˆ’189=09=0f(-\frac{1}{3}) = \frac{-1 - 2 + 21 - 18}{9} = \frac{0}{9} = 0

Alright, so it checks out! Now we can proceed to find the other factors using polynomial division or synthetic division. Knowing one factor is a great head start, making the whole process much smoother. We know (3x+1)(3x + 1) divides f(x)f(x) evenly, so let's use polynomial long division to find the remaining quadratic factor. This method helps us break down the cubic polynomial into simpler terms that are easier to handle. By identifying the first root, we've essentially taken a significant step towards fully factorizing the given polynomial. This initial step confirms our foundation and directs our strategy effectively. Understanding the factor theorem here is crucial as it directly links the roots of the polynomial to its factors. This principle simplifies the factorization process and enables us to proceed with confidence. Recognizing that x=βˆ’13x = -\frac{1}{3} is a root not only validates our approach but also offers a clear path forward in breaking down the cubic equation. Without this crucial first step, the entire factorization process would be far more complex and less intuitive. So, grasping and applying the factor theorem here is not just helpful; it's fundamental to efficiently solving this problem. The importance of this step cannot be overstated, it allows us to transition from a complex cubic polynomial to a simpler quadratic form, setting the stage for easier solutions. In essence, correctly using the factor theorem here transforms a potentially daunting problem into a manageable task. Understanding how polynomials behave around their roots will make you see math in a new light!

Polynomial Long Division

Let's perform polynomial long division to divide f(x)=3x3βˆ’2x2βˆ’7xβˆ’2f(x) = 3x^3 - 2x^2 - 7x - 2 by (3x+1)(3x + 1).

            x^2  - x - 2
      ______________________
3x + 1 | 3x^3 - 2x^2 - 7x - 2
       - (3x^3 + x^2)
       ______________________
             -3x^2 - 7x
             - (-3x^2 - x)
             ______________________
                   -6x - 2
                   - (-6x - 2)
                   ______________________
                         0

So, 3x3βˆ’2x2βˆ’7xβˆ’2=(3x+1)(x2βˆ’xβˆ’2)3x^3 - 2x^2 - 7x - 2 = (3x + 1)(x^2 - x - 2).

Now we need to factorize the quadratic x2βˆ’xβˆ’2x^2 - x - 2. We are looking for two numbers that multiply to -2 and add to -1. Those numbers are -2 and 1. Therefore:

x2βˆ’xβˆ’2=(xβˆ’2)(x+1)x^2 - x - 2 = (x - 2)(x + 1)

Thus, f(x)=(3x+1)(xβˆ’2)(x+1)f(x) = (3x + 1)(x - 2)(x + 1). Performing polynomial long division allows us to systematically reduce the cubic polynomial to a product of simpler factors. This method ensures accuracy and helps in breaking down complex expressions step by step. The meticulous process of dividing each term and subtracting the appropriate multiples reveals the underlying quadratic factor, which in turn can be further factorized. This stage is critical as it transforms the problem from dealing with a cubic to managing a much simpler quadratic equation. The structured approach of long division not only provides the quotient but also confirms that there is no remainder, validating our initial assumption that (3x+1)(3x + 1) is indeed a factor. By carefully aligning terms and following the division algorithm, we minimize errors and gain confidence in our solution. This method essentially unwraps the polynomial, exposing its fundamental components. Mastering this technique is invaluable for any student studying algebra, as it lays the groundwork for handling more advanced polynomial manipulations. Ultimately, polynomial long division serves as a powerful tool for simplifying and solving complex equations, turning what seems like a daunting task into a manageable series of steps. This skill is not just about finding the right answer; it's about understanding the structure and behavior of polynomials, which is essential for further studies in mathematics. Also, be sure to double-check your work because a small error in any step can lead to an incorrect factorization, so attention to detail is key!

Final Factorization

Therefore, the complete factorization of f(x)f(x) is:

f(x)=(3x+1)(xβˆ’2)(x+1)f(x) = (3x + 1)(x - 2)(x + 1)

And there you have it! We've successfully factorized the cubic polynomial f(x)=3x3βˆ’2x2βˆ’7xβˆ’2f(x) = 3x^3 - 2x^2 - 7x - 2 completely, using the given factor (3x+1)(3x + 1) and polynomial division. The final factorization f(x)=(3x+1)(xβˆ’2)(x+1)f(x) = (3x + 1)(x - 2)(x + 1) is the pinnacle of our efforts. This result confirms the roots of the polynomial as x=βˆ’13x = -\frac{1}{3}, x=2x = 2, and x=βˆ’1x = -1. Completing the factorization provides a comprehensive understanding of the polynomial's structure and behavior. Each factor represents a linear component, and their product reconstructs the original cubic equation. This process highlights the fundamental relationship between roots and factors, showcasing how identifying one factor can unlock the entire factorization. Understanding the importance of completing the factorization cannot be understated, as it enables further analysis such as finding roots, graphing the polynomial, and solving related equations. The ability to break down complex polynomials into simpler forms is a cornerstone of algebraic manipulation and is vital for more advanced mathematical studies. By successfully reaching the final factorization, we demonstrate a mastery of the techniques and concepts involved, reinforcing our problem-solving skills. Remember, the journey of factorization involves not just finding the answer, but also understanding the underlying principles and strategies that make it possible. So, keep practicing and expanding your knowledge, and you'll become proficient in handling even the most challenging polynomial problems! Always double check your answer by expanding the factors to make sure it equals the original polynomial. Also, always remember that practice makes perfect!

Conclusion

In summary, factorizing polynomials can seem daunting at first, but with the right toolsβ€”like the factor theorem and polynomial divisionβ€”it becomes a manageable task. Keep practicing, and you'll become a pro in no time! This detailed step-by-step approach to factorizing the cubic polynomial f(x)=3x3βˆ’2x2βˆ’7xβˆ’2f(x) = 3x^3 - 2x^2 - 7x - 2 not only provides the solution but also enhances understanding of the underlying mathematical principles. By utilizing the given factor (3x+1)(3x + 1), we efficiently navigated through polynomial long division to uncover the quadratic factor, which was subsequently factorized into linear components. This process highlights the interconnectedness of roots and factors, and the importance of a systematic approach in simplifying complex algebraic expressions. The final factorization, f(x)=(3x+1)(xβˆ’2)(x+1)f(x) = (3x + 1)(x - 2)(x + 1), offers a complete representation of the polynomial's structure, enabling further analysis and applications. Mastering techniques like polynomial long division and recognizing patterns in quadratic expressions are essential skills for anyone studying algebra. This exercise underscores the value of practice and attention to detail in mathematical problem-solving. By consistently applying these strategies, students can build confidence and proficiency in handling various polynomial factorization problems. Always remember that even complex problems can be broken down into manageable steps, making them more accessible and less intimidating. So keep honing your skills, and you'll become adept at tackling any factorization challenge that comes your way!