Factoring F(x) = X³ - 2x² - 5x + 6 A Step-by-Step Guide

Factoring polynomials is a fundamental skill in algebra, and it's especially important when dealing with cubic polynomials like the one we have here: f(x) = x³ - 2x² - 5x + 6. Finding the completely factored form allows us to easily identify the roots (or zeros) of the polynomial, which are the values of x that make f(x) = 0. This factored form is also crucial for solving polynomial equations, sketching graphs, and simplifying rational expressions. This article will walk you through a detailed process of factoring this specific cubic polynomial, explaining the underlying concepts and techniques along the way. We'll explore the Rational Root Theorem, synthetic division, and quadratic factoring, providing you with a comprehensive understanding of how to approach such problems. Understanding these methods is not just about solving this specific problem; it's about building a strong foundation for more advanced algebraic concepts. Polynomial factorization is a cornerstone of mathematical analysis, finding applications in calculus, differential equations, and various other fields. The ability to efficiently and accurately factor polynomials unlocks the door to solving a wide range of mathematical problems, making it an indispensable skill for students and professionals alike. The process might seem daunting at first, but by breaking it down into manageable steps, we can systematically unravel the polynomial and reveal its factored form. So, let's dive in and explore the fascinating world of polynomial factorization!

1. The Rational Root Theorem: Your First Clue

When faced with a polynomial like f(x) = x³ - 2x² - 5x + 6, the first step in factoring is often to employ the Rational Root Theorem. This powerful theorem provides us with a list of potential rational roots – that is, roots that can be expressed as fractions (including integers). It doesn't guarantee that any of these potential roots are actual roots, but it significantly narrows down the possibilities, giving us a starting point for our investigation. The Rational Root Theorem states that if a polynomial with integer coefficients has rational roots, they must be of the form p/q, where p is a factor of the constant term (the term without any x) and q is a factor of the leading coefficient (the coefficient of the highest power of x). In our case, the constant term is 6, and the leading coefficient is 1. Therefore, the factors of the constant term (p) are ±1, ±2, ±3, and ±6. The factors of the leading coefficient (q) are ±1. This means our potential rational roots (p/q) are ±1, ±2, ±3, and ±6. We now have a manageable list of numbers to test. To determine if any of these are actual roots, we can substitute them into the polynomial f(x). If f(a) = 0 for some value a, then a is a root of the polynomial, and (x - a) is a factor. This is the essence of the Factor Theorem, which is closely related to the Rational Root Theorem. The beauty of the Rational Root Theorem lies in its ability to transform an seemingly intractable problem into a series of straightforward tests. Instead of blindly guessing roots, we have a systematic method for identifying potential candidates. This not only saves time but also provides a structured approach to problem-solving, which is a valuable skill in mathematics and beyond. Remember, this is just the first step, but it's a crucial one in unlocking the factored form of the polynomial.

2. Testing Potential Roots: Synthetic Division to the Rescue

Now that we have our list of potential rational roots (±1, ±2, ±3, ±6) from the Rational Root Theorem, the next step is to test these values to see if any of them are actual roots of the polynomial f(x) = x³ - 2x² - 5x + 6. One efficient method for testing these values is synthetic division. Synthetic division is a streamlined process for dividing a polynomial by a linear factor of the form (x - a). It's a faster and more compact alternative to long division, especially when dealing with polynomials of higher degrees. Let's start by testing the simplest potential root, x = 1. Setting up the synthetic division, we write down the coefficients of the polynomial (1, -2, -5, 6) and the test value (1) as follows:

1 | 1 -2 -5 6
  |________________

We bring down the first coefficient (1), multiply it by the test value (1), and write the result under the next coefficient (-2). Then, we add the two numbers (-2 and 1) and write the sum (-1) below. We repeat this process: multiply -1 by 1, write the result (-1) under -5, add them (-5 and -1) to get -6, multiply -6 by 1, write -6 under 6, and finally add 6 and -6 to get 0. The final result looks like this:

1 | 1 -2 -5 6
  |   1 -1 -6
  |________________
    1 -1 -6 0

The last number in the bottom row (0) is the remainder. If the remainder is 0, then the test value (1) is a root of the polynomial, and (x - 1) is a factor. In this case, the remainder is 0, so x = 1 is indeed a root. The other numbers in the bottom row (1, -1, -6) are the coefficients of the quotient polynomial, which is one degree lower than the original polynomial. Therefore, the quotient polynomial is x² - x - 6. Synthetic division not only tells us if a value is a root but also gives us the quotient polynomial, which is a crucial piece of information for further factoring. We've successfully found one factor of our cubic polynomial! Now we can focus on factoring the resulting quadratic. But before we move on, let's emphasize the power of synthetic division. It's a workhorse technique for polynomial factorization, allowing us to quickly test potential roots and reduce the complexity of the problem.

3. Factoring the Quadratic: Completing the Puzzle

Having used synthetic division to determine that (x - 1) is a factor of f(x) = x³ - 2x² - 5x + 6, we've also obtained the quotient polynomial, which is x² - x - 6. Now, to find the completely factored form of the original cubic polynomial, we need to factor this quadratic. Factoring quadratics is a more straightforward process compared to cubics. There are several techniques we can use, such as looking for two numbers that multiply to the constant term (-6) and add up to the coefficient of the linear term (-1), or using the quadratic formula. In this case, we can easily find two such numbers: -3 and 2. These numbers satisfy the conditions because (-3) * (2) = -6 and (-3) + (2) = -1. Therefore, we can factor the quadratic as (x - 3)(x + 2). This step is a crucial link in the chain of factorization. We started with a cubic polynomial and, through the Rational Root Theorem and synthetic division, we reduced it to a quadratic. Now, by factoring the quadratic, we're essentially breaking down the polynomial into its most fundamental linear factors. Each linear factor corresponds to a root of the polynomial, giving us a complete understanding of its behavior. It's like dismantling a complex machine into its individual components, allowing us to see how each part contributes to the whole. Factoring the quadratic is often the final step in factoring a cubic polynomial, but it's important to remember that the techniques used here are applicable to a wide range of quadratic expressions. Mastering quadratic factoring is an essential skill in algebra, with applications in solving equations, graphing functions, and simplifying expressions. Now that we've factored the quadratic, we have all the pieces of the puzzle. We know that f(x) can be expressed as the product of three linear factors. The final step is to combine these factors to obtain the completely factored form.

4. The Completely Factored Form: Putting it All Together

We've come a long way in our journey to factor the cubic polynomial f(x) = x³ - 2x² - 5x + 6. We started by using the Rational Root Theorem to identify potential rational roots, then employed synthetic division to confirm that x = 1 is a root and obtained the quotient polynomial x² - x - 6. Finally, we factored the quadratic to get (x - 3)(x + 2). Now, we can combine these pieces to express f(x) in its completely factored form. Since we found that (x - 1) is a factor and x² - x - 6 = (x - 3)(x + 2), we can write the completely factored form of f(x) as:

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

This is the final answer, and it represents the polynomial as a product of linear factors. Each factor corresponds to a root of the polynomial: x = 1, x = 3, and x = -2. These are the values of x that make f(x) = 0. The completely factored form is not just a solution; it's a powerful representation of the polynomial. It allows us to easily visualize the roots, sketch the graph, and analyze the behavior of the function. For example, we can see that the graph of f(x) will cross the x-axis at x = 1, x = 3, and x = -2. The factored form also makes it easier to solve equations involving the polynomial. If we want to find the solutions to f(x) = 0, we simply set each factor equal to zero and solve for x. This gives us the same roots we identified earlier. In conclusion, finding the completely factored form of a polynomial is a fundamental skill in algebra, with wide-ranging applications. By understanding and applying techniques like the Rational Root Theorem, synthetic division, and quadratic factoring, we can systematically break down complex polynomials into their simplest components. This not only solves the immediate problem but also provides a deeper understanding of the polynomial's structure and behavior. The journey of factoring this cubic polynomial has been a testament to the power of algebraic techniques and their ability to reveal the hidden structure within mathematical expressions.

Therefore, the correct answer is:

D. f(x) = (x + 2)(x - 3)(x - 1)