Factoring Polynomials Unveiling The Completely Factored Form Of F(x)=6x³-13x²-4x+15
Hey guys! Today, we're diving deep into the world of polynomial factorization. Specifically, we're going to tackle the question: What is the completely factored form of the cubic polynomial f(x) = 6x³ - 13x² - 4x + 15? This type of problem might seem daunting at first, but with a systematic approach and a few key techniques, we can break it down and find the solution. Polynomial factorization is a fundamental concept in algebra, and mastering it is crucial for solving higher-level mathematical problems. Whether you're a student preparing for an exam or just someone looking to brush up on your math skills, this guide will walk you through the process step-by-step. We'll explore the rational root theorem, synthetic division, and quadratic factorization to arrive at the completely factored form of the given polynomial. So, let's get started and unravel this mathematical puzzle together!
Cracking the Code: Factoring Polynomials Like a Pro
To find the completely factored form of f(x) = 6x³ - 13x² - 4x + 15, we'll employ a combination of algebraic techniques. First off, the journey of factoring polynomials, especially cubic ones like our f(x), often begins with a bit of educated guesswork and some clever tools. Our main mission here is to break down this complex expression into simpler, more manageable pieces – think of it like disassembling a complicated machine into its individual components. These components, in the world of polynomials, are usually linear factors (expressions like x - a) and sometimes quadratic factors (expressions like ax² + bx + c). When we multiply these factors together, they should give us back our original polynomial. This process isn't just a mathematical exercise; it's a fundamental skill in algebra with wide-ranging applications, from solving equations to simplifying complex expressions. Factoring polynomials allows us to understand the behavior of polynomial functions, find their roots (the x-intercepts on a graph), and solve real-world problems involving rates, areas, and volumes. So, with our goal clearly defined, let's dive into the methods we can use to achieve this factorization.
Step 1: The Rational Root Theorem – Our Detective's Toolkit
The Rational Root Theorem is our first big weapon in this factoring quest. This theorem is like a detective's toolkit, giving us a list of potential suspects – possible rational roots of our polynomial. A rational root is simply a root (a value of x that makes the polynomial equal to zero) that can be expressed as a fraction (p/q). The theorem states that if a polynomial has integer coefficients, then any rational root must be of the form ±p/q, where p is a factor of the constant term (the term without any x – in our case, 15) and q is a factor of the leading coefficient (the coefficient of the highest power of x – in our case, 6). So, let's put on our detective hats and gather our suspects. The factors of 15 are ±1, ±3, ±5, and ±15. The factors of 6 are ±1, ±2, ±3, and ±6. Now, we form all possible fractions ±p/q using these factors. This gives us a list of potential rational roots: ±1, ±1/2, ±1/3, ±1/6, ±3, ±3/2, ±5, ±5/2, ±5/3, ±5/6, ±15, and ±15/2. Woah, that's quite a lineup! But don't worry, we won't have to try all of them. This list just gives us a starting point for our investigation. We've narrowed down the possibilities significantly, and now we're ready to start testing these potential roots to see if any of them actually make our polynomial equal to zero. This is where the next step, synthetic division, comes into play.
Step 2: Synthetic Division – Testing the Suspects
Now that we have our list of potential rational roots, it's time to put them to the test. Synthetic division is our method of choice here. Think of it as a quick and efficient way to check if a particular number is a root of the polynomial and, if it is, to also find the quotient when the polynomial is divided by (x - root). It's like a magic trick that simplifies the process of polynomial division. Let's start with an easy suspect from our list – say, x = 1. We set up our synthetic division table with the coefficients of our polynomial (6, -13, -4, 15) and the potential root (1). The process involves bringing down the first coefficient, multiplying it by the root, adding it to the next coefficient, and repeating this process until we reach the end. If the final result (the remainder) is zero, then we've found a root! If the remainder is not zero, then that number is not a root, and we move on to the next suspect. Let's try it out with x = -1. Performing synthetic division with -1, we get a remainder of 0. This means that x = -1 is indeed a root of our polynomial, and (x + 1) is a factor. This is a major breakthrough! Finding one factor is like finding the first piece of a puzzle – it opens up the way to finding the rest. But the beauty of synthetic division doesn't stop there. It also gives us the coefficients of the quotient polynomial, which is one degree lower than our original polynomial. In this case, the quotient is a quadratic, which we can then try to factor further using techniques like factoring by grouping or the quadratic formula. So, by using synthetic division, we've not only identified a root but also simplified our polynomial into a more manageable form. Now, let's see what the quotient polynomial looks like and how we can factor it further.
Step 3: Unveiling the Quadratic Quotient
After performing synthetic division with x = -1, we've discovered that it is indeed a root and that (x + 1) is a factor of our polynomial f(x) = 6x³ - 13x² - 4x + 15. But the magic of synthetic division goes beyond just identifying a root. It also hands us the quotient polynomial on a silver platter. The numbers we get in the bottom row of our synthetic division (excluding the remainder, which is 0) are the coefficients of this quotient polynomial. In our case, these coefficients are 6, -19, and 15. This means that when we divide our original cubic polynomial by (x + 1), we get a quadratic polynomial: 6x² - 19x + 15. This is a huge step forward! We've effectively reduced a cubic polynomial factorization problem into a quadratic polynomial factorization problem. Quadratics are generally much easier to handle. We have a variety of tools at our disposal, such as factoring by grouping, using the quadratic formula, or simply trying to find two numbers that multiply to give the constant term and add up to give the coefficient of the linear term. The quadratic polynomial 6x² - 19x + 15 is now our focus. Our goal is to break it down into two linear factors, which will complete the factorization of our original cubic polynomial. So, let's put on our quadratic factorization hats and see how we can crack this one. There are several paths we can take, and we'll explore the most common and effective ones to find the factors of this quadratic.
Step 4: Factoring the Quadratic – The Final Showdown
Now we've arrived at the crucial step: factoring the quadratic 6x² - 19x + 15. There are a couple of ways we can approach this. One common method is to look for two numbers that multiply to give the product of the leading coefficient (6) and the constant term (15), which is 90, and add up to the middle coefficient (-19). This might sound a bit like a puzzle, but with a little trial and error, we can find those numbers. In this case, the numbers are -10 and -9 because (-10) * (-9) = 90 and (-10) + (-9) = -19. Once we've found these numbers, we can rewrite the middle term of the quadratic (-19x) as the sum of these two numbers multiplied by x: -10x - 9x. This allows us to factor by grouping. We rewrite the quadratic as 6x² - 10x - 9x + 15. Now, we group the first two terms and the last two terms: (6x² - 10x) + (-9x + 15). We can factor out the greatest common factor (GCF) from each group. From the first group, we can factor out 2x, which gives us 2x(3x - 5). From the second group, we can factor out -3, which gives us -3(3x - 5). Notice that we now have a common factor of (3x - 5) in both terms. We can factor this out, leaving us with (2x - 3)(3x - 5). And there you have it! We've successfully factored the quadratic 6x² - 19x + 15 into (2x - 3)(3x - 5). This is the final piece of the puzzle. We've broken down the quadratic into its linear factors, and now we're just one step away from the completely factored form of our original cubic polynomial. Let's bring it all together in the final step.
Step 5: The Grand Finale – Putting It All Together
We've reached the climax of our factoring adventure! We've successfully navigated through the Rational Root Theorem, conquered synthetic division, and triumphed in factoring the quadratic. Now, it's time to assemble all the pieces and reveal the completely factored form of our polynomial f(x) = 6x³ - 13x² - 4x + 15. Remember, we started by using the Rational Root Theorem to identify potential rational roots. Then, we employed synthetic division and found that x = -1 is a root, which means (x + 1) is a factor. The synthetic division also gave us the quotient polynomial, 6x² - 19x + 15. We then factored this quadratic into (2x - 3)(3x - 5) using factoring by grouping. Now, we simply combine all these factors together. Our original polynomial f(x) can be expressed as the product of (x + 1) and the factored quadratic (2x - 3)(3x - 5). Therefore, the completely factored form of f(x) = 6x³ - 13x² - 4x + 15 is (x + 1)(2x - 3)(3x - 5). That's it! We've cracked the code and found the completely factored form. This is a significant accomplishment. We've not only solved this specific problem but also honed our skills in polynomial factorization, a valuable tool in algebra and beyond. Factoring polynomials allows us to solve equations, simplify expressions, and understand the behavior of polynomial functions. So, give yourselves a pat on the back for sticking with it and mastering this important concept!
The Ultimate Answer
Therefore, the completely factored form of f(x) = 6x³ - 13x² - 4x + 15 is:
(x + 1)(2x - 3)(3x - 5)
So, there you have it, guys! We've successfully navigated the world of polynomial factorization and found the completely factored form of our cubic polynomial. This journey wasn't just about finding the answer; it was about learning the process, understanding the techniques, and building our problem-solving skills. Polynomial factorization is a fundamental concept in algebra, and it's a skill that will serve you well in more advanced math courses and beyond. Whether you're tackling calculus, differential equations, or even real-world applications of mathematics, the ability to factor polynomials will be a valuable asset. Remember, practice makes perfect. The more you work with these techniques, the more comfortable and confident you'll become. Don't be afraid to tackle challenging problems and explore different approaches. The beauty of mathematics lies in its ability to reveal patterns and connections, and the more you explore, the more you'll discover. So, keep practicing, keep learning, and keep pushing your mathematical boundaries. And who knows, maybe one day you'll be the one unlocking the secrets of even more complex mathematical puzzles!
