Factoring Polynomials Remainder Theorem Find All Factors

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Polynomial factorization is a fundamental concept in algebra, playing a crucial role in solving equations, simplifying expressions, and understanding the behavior of polynomial functions. In this comprehensive guide, we will delve into the process of factoring polynomials, with a specific focus on utilizing the Remainder Theorem. Our goal is to equip you with the knowledge and skills necessary to confidently tackle polynomial factorization problems. This article will help you understand the nuances of factoring, especially when a root is already known. We will use the Remainder Theorem as our guide, ensuring that you grasp not just the how but also the why behind each step. Let's embark on this journey together, transforming complex equations into manageable components, and unlocking the secrets held within polynomial expressions. Remember, mastering polynomial factorization opens doors to advanced mathematical concepts and real-world applications. So, whether you are a student grappling with algebra or someone seeking a refresher, this guide will provide you with the tools you need to succeed.

Before we dive into the solution, let's clearly define the problem at hand. We are given a cubic polynomial function, denoted as f(x) = x³ - 4x² - 20x + 48. Our primary task is to factor this polynomial completely, which means expressing it as a product of its linear factors. We are also provided with a crucial piece of information: one root of the function is x = 6. This root serves as our starting point, acting as a key to unlock the complete factorization. Understanding the given polynomial and the significance of the known root is essential. The root x = 6 tells us that (x - 6) is one of the factors of the polynomial. The challenge now is to find the remaining factors. Factoring polynomials is akin to solving a puzzle, and each piece of information is a clue. In this case, the Remainder Theorem will be our guiding principle, helping us systematically unravel the factors of f(x). Remember, polynomials are the building blocks of many mathematical models, and the ability to factor them is a powerful skill.

The Remainder Theorem is a cornerstone of polynomial algebra, providing a powerful connection between polynomial division and the roots of a polynomial. In essence, the Remainder Theorem states that if a polynomial f(x) is divided by (x - c), then the remainder is f(c). This theorem is particularly useful in our context because it allows us to check if a given value is a root of the polynomial. If f(c) = 0, then (x - c) is a factor of f(x). In our case, we are given that x = 6 is a root. This means that f(6) should equal zero, and (x - 6) is a factor of our polynomial. The Remainder Theorem not only helps us verify roots but also provides a systematic way to find other factors. By dividing the original polynomial by a known factor, we can reduce the degree of the polynomial, making it easier to factor further. Understanding the Remainder Theorem is crucial for efficiently factoring polynomials, especially when combined with techniques like synthetic division or polynomial long division. It's a fundamental tool in the algebraist's toolkit, bridging the gap between polynomial division and root finding.

Now, let's put the Remainder Theorem into action. We know that x = 6 is a root of the polynomial f(x) = x³ - 4x² - 20x + 48. This implies that (x - 6) is a factor. To find the remaining factors, we can divide the polynomial f(x) by (x - 6). We will employ polynomial long division for this purpose. The process involves dividing the highest degree term of the dividend (x³) by the highest degree term of the divisor (x), which gives us x². We then multiply (x - 6) by x² and subtract the result from the original polynomial. This process is repeated until we obtain a remainder. In this case, the division yields x² + 2x - 8. This is the quotient, and since x = 6 is a root, the remainder should be zero, confirming that (x - 6) is indeed a factor. The quotient, x² + 2x - 8, is a quadratic polynomial, which is much easier to factor than the original cubic polynomial. Applying the Remainder Theorem and polynomial division has effectively simplified our problem, leading us closer to the complete factorization.

To better illustrate the process, let's walk through the polynomial long division step-by-step. We're dividing x³ - 4x² - 20x + 48 by (x - 6).

  1. Divide the first term of the dividend (x³) by the first term of the divisor (x), which gives x². Write x² above the x² term in the dividend.
  2. Multiply the divisor (x - 6) by x², resulting in x³ - 6x². Write this below the corresponding terms in the dividend.
  3. Subtract the result from the dividend: (x³ - 4x²) - (x³ - 6x²) = 2x². Bring down the next term from the dividend, -20x, resulting in 2x² - 20x.
  4. Divide the first term of the new expression (2x²) by the first term of the divisor (x), which gives 2x. Write +2x above the x term in the dividend.
  5. Multiply the divisor (x - 6) by 2x, resulting in 2x² - 12x. Write this below the 2x² - 20x.
  6. Subtract the result: (2x² - 20x) - (2x² - 12x) = -8x. Bring down the last term from the dividend, +48, resulting in -8x + 48.
  7. Divide the first term of the new expression (-8x) by the first term of the divisor (x), which gives -8. Write -8 above the constant term in the dividend.
  8. Multiply the divisor (x - 6) by -8, resulting in -8x + 48. Write this below the -8x + 48.
  9. Subtract the result: (-8x + 48) - (-8x + 48) = 0. The remainder is 0.

This process confirms that the quotient is x² + 2x - 8, and the remainder is 0, as expected. This detailed walkthrough provides a clear understanding of how polynomial long division works in practice.

After performing polynomial long division, we obtained the quadratic quotient x² + 2x - 8. Our next step is to factor this quadratic expression. Factoring a quadratic involves finding two binomials that, when multiplied together, give us the original quadratic. In this case, we are looking for two numbers that multiply to -8 and add up to 2. By considering the factors of -8, we can identify that 4 and -2 satisfy these conditions (4 * -2 = -8 and 4 + (-2) = 2). Therefore, the quadratic x² + 2x - 8 can be factored as (x + 4)(x - 2). This step is crucial because it breaks down the quadratic into its linear factors, which are the building blocks of the polynomial. Factoring the quadratic quotient completes the factorization process, allowing us to express the original cubic polynomial as a product of linear factors. Remember, practice is key to mastering quadratic factorization, and recognizing patterns in numbers will significantly speed up the process.

Now that we have factored the quadratic quotient, we can write the complete factorization of the original polynomial. We started with f(x) = x³ - 4x² - 20x + 48 and found that (x - 6) is a factor. After dividing f(x) by (x - 6), we obtained the quotient x² + 2x - 8. We then factored this quadratic into (x + 4)(x - 2). Therefore, the complete factorization of f(x) is (x - 6)(x + 4)(x - 2). This means that the polynomial f(x) can be expressed as the product of these three linear factors. The roots of the polynomial are the values of x that make each factor equal to zero, which are x = 6, x = -4, and x = 2. The complete factorization provides a comprehensive understanding of the polynomial's structure and its roots. It's the final step in our journey, showcasing the power of the Remainder Theorem and polynomial division in unraveling complex polynomial expressions. This process not only helps in solving equations but also in understanding the behavior and properties of polynomial functions.

To ensure the accuracy of our factorization, it's always a good practice to verify the result. We can do this by multiplying the factors we obtained and checking if they indeed give us the original polynomial. Our factors are (x - 6), (x + 4), and (x - 2). Let's multiply them together:

First, multiply (x + 4) and (x - 2):

(x + 4)(x - 2) = x² - 2x + 4x - 8 = x² + 2x - 8

Now, multiply the result by (x - 6):

(x - 6)(x² + 2x - 8) = x(x² + 2x - 8) - 6(x² + 2x - 8) = x³ + 2x² - 8x - 6x² - 12x + 48 = x³ - 4x² - 20x + 48

The result of the multiplication is x³ - 4x² - 20x + 48, which is exactly our original polynomial f(x). This verification confirms that our factorization is correct. Verifying the factors is a crucial step in the problem-solving process, as it helps catch any potential errors and ensures confidence in the final answer. It reinforces the understanding of the relationship between factors and the original polynomial.

In this comprehensive guide, we have successfully factored the cubic polynomial f(x) = x³ - 4x² - 20x + 48 using the Remainder Theorem and polynomial long division. We started by understanding the problem, recognizing the importance of the given root x = 6. We then reviewed the Remainder Theorem and applied it to verify that (x - 6) is indeed a factor. Polynomial long division allowed us to divide the original polynomial by (x - 6), resulting in a quadratic quotient. We factored this quadratic quotient and, finally, pieced together the complete factorization of f(x) as (x - 6)(x + 4)(x - 2). We also emphasized the importance of verifying the factors to ensure accuracy. Mastering polynomial factorization is a valuable skill in algebra and beyond. It not only helps in solving equations but also provides a deeper understanding of polynomial functions and their properties. With practice and a solid understanding of the Remainder Theorem, polynomial division, and quadratic factoring techniques, you can confidently tackle a wide range of factorization problems. Remember, the key is to approach each problem systematically, breaking it down into smaller, manageable steps.

Final Answer: The final answer is (x−2)(x+4)(x−6)\boxed{(x-2)(x+4)(x-6)}