Finding Potential Roots Of Polynomial P(x) = X^4 - 9x^2 - 4x + 12
Hey guys! Let's dive into a fun math problem today. We're going to figure out the potential roots of the polynomial p(x) = x^4 - 9x^2 - 4x + 12. Sounds like a mouthful, right? Don't worry, we'll break it down step by step. We have a bunch of options to choose from: A. 0, B. ±2, C. ±4, D. ±9, E. 1/2, F. ±3, G. ±6, H. ±12. So, how do we even start to tackle this? Well, let's get to it and make math feel less like a puzzle and more like a game!
Understanding the Rational Root Theorem
Okay, first things first, let's talk about the Rational Root Theorem. This is our main tool for finding potential rational roots of a polynomial. Basically, this theorem tells us that if a polynomial has rational roots (roots that can be expressed as a fraction), they're going to be hiding out among the factors of the constant term (the number without an 'x') divided by the factors of the leading coefficient (the number in front of the highest power of 'x').
In our case, the polynomial is p(x) = x^4 - 9x^2 - 4x + 12. So, what's the constant term? That's right, it's 12. And the leading coefficient? It's 1 (since there's no number explicitly written in front of x^4, we assume it's 1). Now, let's list out the factors of these numbers. The factors of 12 are ±1, ±2, ±3, ±4, ±6, and ±12. The factors of 1 are simply ±1. According to the Rational Root Theorem, any rational roots of our polynomial must be in the form of (factors of 12) / (factors of 1), which gives us ±1, ±2, ±3, ±4, ±6, and ±12. These are our potential rational roots – the ones we need to check. So, the Rational Root Theorem is super useful because it narrows down the possibilities significantly. Instead of guessing from an infinite number of options, we have a manageable list to work with. Remember, the theorem doesn't guarantee that these numbers are roots, but it tells us where to look first. It’s like having a treasure map – it shows you where the treasure might be, but you still have to do some digging!
Applying the Rational Root Theorem to Our Polynomial
Alright, now that we've got our list of potential rational roots thanks to the Rational Root Theorem, which include ±1, ±2, ±3, ±4, ±6, and ±12, it's time to put them to the test. The easiest way to check if a number is a root of the polynomial is to plug it in and see if we get zero. If p(a) = 0, then 'a' is a root. This is a direct application of the definition of a root – a value that makes the polynomial equal to zero. Let's start with the simpler numbers first, because, why not make life a little easier? We'll try 2 and -2.
Let's plug in 2 into p(x) = x^4 - 9x^2 - 4x + 12: p(2) = (2)^4 - 9(2)^2 - 4(2) + 12 = 16 - 36 - 8 + 12 = -16. So, 2 is not a root. Okay, let's try -2: p(-2) = (-2)^4 - 9(-2)^2 - 4(-2) + 12 = 16 - 36 + 8 + 12 = 0. Bingo! -2 is a root. This is great news! Finding one root helps us to simplify the polynomial, which we'll get to later. Now, let's check out some other options. How about 3? p(3) = (3)^4 - 9(3)^2 - 4(3) + 12 = 81 - 81 - 12 + 12 = 0. Awesome, 3 is also a root! So, by directly substituting the potential roots into the polynomial, we can quickly identify which ones actually make the polynomial equal to zero. This method is straightforward and effective, especially when combined with the Rational Root Theorem, which helps us narrow down our choices. Remember, each time we find a root, we're one step closer to fully understanding our polynomial and its behavior. It’s like solving a puzzle, where each root is a piece that fits into place.
Testing the Potential Roots
Now, let's systematically test the potential roots we identified earlier, which are ±1, ±2, ±3, ±4, ±6, and ±12. We already found that -2 and 3 are roots. Let's continue checking the other options. We'll start with -3.
Let’s calculate p(-3): p(-3) = (-3)^4 - 9(-3)^2 - 4(-3) + 12 = 81 - 81 + 12 + 12 = 24. So, -3 is not a root. Next, let’s try 4: p(4) = (4)^4 - 9(4)^2 - 4(4) + 12 = 256 - 144 - 16 + 12 = 108. So, 4 is not a root either. Now, let’s check -4: p(-4) = (-4)^4 - 9(-4)^2 - 4(-4) + 12 = 256 - 144 + 16 + 12 = 140. Again, -4 is not a root. Let's move on to 6: p(6) = (6)^4 - 9(6)^2 - 4(6) + 12 = 1296 - 324 - 24 + 12 = 960. Clearly, 6 is not a root. Now, let’s try -6: p(-6) = (-6)^4 - 9(-6)^2 - 4(-6) + 12 = 1296 - 324 + 24 + 12 = 1008. So, -6 is not a root. Lastly, let's check 12: p(12) = (12)^4 - 9(12)^2 - 4(12) + 12 = 20736 - 1296 - 48 + 12 = 19404. Obviously, 12 is not a root. And finally, -12: p(-12) = (-12)^4 - 9(-12)^2 - 4(-12) + 12 = 20736 - 1296 + 48 + 12 = 19500. So, -12 is not a root either. After testing all the potential roots, we found that only -2 and 3 are actual roots of the polynomial. This process highlights the importance of systematic testing. While the Rational Root Theorem gives us a manageable list of possibilities, we still need to check each one to confirm whether it's a root. It might seem tedious, but it’s a crucial step in solving polynomial equations and understanding their behavior. Each test gives us more information and brings us closer to the complete solution. It’s like being a detective, where every clue helps you solve the case!
Using Synthetic Division to Simplify the Polynomial
Okay, so far we've identified two roots of our polynomial p(x) = x^4 - 9x^2 - 4x + 12: -2 and 3. That’s a solid start! Now, let's leverage these roots to make our polynomial simpler. We're going to use a technique called synthetic division. Synthetic division is a super handy shortcut for dividing a polynomial by a linear factor (like x - a, where 'a' is a root). It's much faster and cleaner than long division, especially for higher-degree polynomials. Think of it as the express lane for polynomial division!
First, let's use synthetic division with the root -2. We set up the synthetic division table using the coefficients of our polynomial (1, 0, -9, -4, 12). Note that we include a 0 for the missing x^3 term. The setup looks something like this:
-2 | 1 0 -9 -4 12
|____________________
Now, we perform the synthetic division steps. We bring down the first coefficient (1), multiply it by -2, and write the result under the next coefficient (0). We add those numbers, and repeat the process:
-2 | 1 0 -9 -4 12
| -2 4 10 -12
|____________________
1 -2 -5 6 0
The last number in the bottom row is 0, which confirms that -2 is indeed a root (the remainder is 0). The other numbers (1, -2, -5, 6) are the coefficients of the quotient polynomial, which is x^3 - 2x^2 - 5x + 6. So, dividing p(x) by (x + 2) gives us x^3 - 2x^2 - 5x + 6. This is awesome because we've reduced the degree of our polynomial from 4 to 3, making it easier to handle. Now, let's use synthetic division again, this time with the root 3, on our new polynomial x^3 - 2x^2 - 5x + 6:
3 | 1 -2 -5 6
| 3 3 -6
|___________
1 1 -2 0
Again, the remainder is 0, confirming that 3 is a root of x^3 - 2x^2 - 5x + 6. The quotient polynomial is now x^2 + x - 2. We've successfully reduced our original fourth-degree polynomial to a quadratic polynomial! Synthetic division has been a game-changer here. By using it twice, we've simplified the problem significantly. Now, we’re dealing with a quadratic, which we can solve using factoring, the quadratic formula, or completing the square. This is a perfect example of how breaking down a complex problem into smaller, manageable steps can make it much easier to solve. Each division brings us closer to finding all the roots and fully understanding the polynomial’s structure. It’s like peeling back the layers of an onion, where each layer reveals more about the core.
Solving the Remaining Quadratic Equation
Alright, we've made some serious progress! We started with the polynomial p(x) = x^4 - 9x^2 - 4x + 12, found two roots (-2 and 3), and used synthetic division to reduce it down to the quadratic equation x^2 + x - 2 = 0. This is fantastic because solving quadratic equations is something we can do with ease. There are several methods we can use, such as factoring, using the quadratic formula, or completing the square. In this case, factoring looks like the simplest route.
We need to find two numbers that multiply to -2 and add up to 1 (the coefficient of the x term). Those numbers are 2 and -1. So, we can factor the quadratic as (x + 2)(x - 1) = 0. Now, to find the roots, we set each factor equal to zero:
x + 2 = 0 gives us x = -2
x - 1 = 0 gives us x = 1
So, the roots of the quadratic equation x^2 + x - 2 = 0 are -2 and 1. Now, let's put it all together. We initially found the roots -2 and 3 using the Rational Root Theorem and direct substitution. Then, we used synthetic division to reduce the polynomial to a quadratic, and we found the remaining roots to be -2 and 1. Therefore, the roots of the original polynomial p(x) = x^4 - 9x^2 - 4x + 12 are -2, 3, -2, and 1. Notice that the root -2 appears twice, which means it has a multiplicity of 2. Understanding the roots of a polynomial is incredibly powerful. It allows us to fully factor the polynomial and understand its behavior, such as where it crosses the x-axis. This process of using the Rational Root Theorem, synthetic division, and solving the resulting quadratic equation is a classic strategy for tackling higher-degree polynomials. It’s like having a set of tools in a toolbox – each tool serves a specific purpose, and when used together, they can help you solve even the most challenging problems. It’s a testament to the elegance and power of mathematical techniques!
Final Answer and Reflection
Alright, let's wrap things up and give the final answer to our original question: What are the potential roots of the polynomial p(x) = x^4 - 9x^2 - 4x + 12? We started with a list of options: A. 0, B. ±2, C. ±4, D. ±9, E. 1/2, F. ±3, G. ±6, H. ±12. Through our journey of using the Rational Root Theorem, testing potential roots, and applying synthetic division, we discovered that the actual roots are -2, 3, -2, and 1. So, from the given options, the potential roots that actually turned out to be roots are ±2 and ±3.
Therefore, the correct options are B (±2) and F (±3). Now, let's take a moment to reflect on what we've done. We didn't just blindly guess the answer; we used a systematic approach. We applied the Rational Root Theorem to narrow down the possibilities, tested those possibilities using direct substitution, and then employed synthetic division to simplify the polynomial. Finally, we solved the resulting quadratic equation to find the remaining roots. This step-by-step method is key to tackling polynomial problems, and math problems in general. It's like building a house – you need a solid foundation (the Rational Root Theorem), strong walls (testing potential roots), and a sturdy roof (synthetic division and solving the quadratic). Each step builds upon the previous one, leading us to the final solution. More importantly, we've seen how different mathematical tools can work together to solve a complex problem. The Rational Root Theorem gave us a starting point, synthetic division made the polynomial more manageable, and our knowledge of quadratic equations allowed us to finish the job. It’s a beautiful example of how different areas of math connect and complement each other. So, next time you encounter a polynomial problem, remember this journey. Don't be intimidated by the complexity; break it down into smaller steps, use the right tools, and enjoy the process of discovery. Math isn’t just about getting the right answer; it’s about the journey of solving the problem and the insights you gain along the way. Keep exploring, keep questioning, and keep having fun with math! You guys got this!
