Polynomial Remainder Theorem: Solving For F(x)

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Hey guys! Ever get those math problems that seem like a puzzle? Today, we're diving into one of those, specifically dealing with polynomials and remainders. This type of problem often uses the Remainder Theorem, which is a super handy tool when you're dealing with polynomial division. So, let's break down this polynomial problem step by step and make sure we understand the core concepts. We'll tackle a specific question, but the techniques we learn will be useful for a bunch of similar problems. Let's get started!

Understanding the Problem

Okay, so here's the problem we're going to solve: A polynomial F(x) leaves a remainder of -5 when divided by (x + 2), and it leaves a remainder of 3 when divided by (x - 2). The question is, what's the remainder when F(x) is divided by x² - 4? Sounds a bit tricky, right? But don't worry, we'll make it crystal clear.

First, let's identify the key information. We know two crucial things: When F(x) is divided by (x + 2), the remainder is -5. This means that if we plug in x = -2 into F(x), we should get -5. Mathematically, this is written as F(-2) = -5. Similarly, when F(x) is divided by (x - 2), the remainder is 3. This means that F(2) = 3. These two pieces of information are our foundation for solving the problem.

Now, consider what happens when we divide F(x) by x² - 4. Since x² - 4 is a quadratic (degree 2) polynomial, the remainder will be a polynomial of degree at most 1. In other words, the remainder will be of the form ax + b, where a and b are constants that we need to find. This is a crucial step because it allows us to express the remainder in a manageable form. We're not looking for a complicated polynomial; we're looking for a simple linear expression.

To summarize, we've identified the given information (F(-2) = -5 and F(2) = 3), understood the form of the remainder (ax + b), and set the stage for using the Polynomial Remainder Theorem to solve for a and b. The next steps will involve applying the theorem and solving a system of equations. So far so good, right? Let's move on and put these pieces together!

Applying the Remainder Theorem

Alright, let's get our hands dirty with the Remainder Theorem. This theorem is the backbone of solving this problem, so let's break it down and see how it applies here. The Remainder Theorem basically states that if you divide a polynomial F(x) by (x - c), the remainder is F(c). We've already seen this in action when we stated that F(-2) = -5 and F(2) = 3. But let's see how we can use this to find the remainder when F(x) is divided by x² - 4.

We know that x² - 4 can be factored into (x + 2)(x - 2). This is super helpful because it connects our divisor to the information we already have about the remainders when F(x) is divided by (x + 2) and (x - 2) individually. Now, let's represent the division of F(x) by x² - 4. We can write this as:

F(x) = (x² - 4)Q(x) + R(x)

Where Q(x) is the quotient (the result of the division) and R(x) is the remainder. Remember, we already figured out that the remainder R(x) will be in the form ax + b. So, we can rewrite the equation as:

F(x) = (x² - 4)Q(x) + ax + b

Now comes the clever part. We're going to use the values x = -2 and x = 2 that we know from the problem. Let's plug in x = -2:

F(-2) = ((-2)² - 4)Q(-2) + a(-2) + b

Since F(-2) = -5 and (-2)² - 4 = 0, this simplifies to:

-5 = -2a + b

Great! We've got our first equation. Now, let's plug in x = 2:

F(2) = ((2)² - 4)Q(2) + a(2) + b

Since F(2) = 3 and (2)² - 4 = 0, this simplifies to:

3 = 2a + b

And there we have it! We've got two equations with two unknowns (a and b):

  1. -5 = -2a + b
  2. 3 = 2a + b

We've successfully translated the polynomial problem into a system of linear equations. The Remainder Theorem has allowed us to connect the remainders to the coefficients of the remainder polynomial. In the next section, we'll solve this system of equations to find the values of a and b, which will give us the remainder R(x). Almost there, guys! Let's keep pushing!

Solving for the Remainder

Okay, we've arrived at the fun part – solving the system of equations! We have two equations:

  1. -5 = -2a + b
  2. 3 = 2a + b

There are a couple of ways we can solve this. One popular method is elimination, and that's what we'll use here. Notice that the 2a terms in the two equations have opposite signs. This makes elimination a breeze. If we add the two equations together, the 2a terms will cancel out:

(-5) + (3) = (-2a + b) + (2a + b)

This simplifies to:

-2 = 2b

Now, we can easily solve for b by dividing both sides by 2:

b = -1

Awesome! We've found the value of b. Now we need to find a. We can plug the value of b into either of our original equations. Let's use the second equation, 3 = 2a + b:

3 = 2a + (-1)

Add 1 to both sides:

4 = 2a

Now, divide both sides by 2:

a = 2

Fantastic! We've found both a and b: a = 2 and b = -1. Remember, the remainder R(x) is in the form ax + b. So, we can now write the remainder as:

R(x) = 2x - 1

And that's it! We've successfully found the remainder when F(x) is divided by x² - 4. It's 2x - 1. This matches option a in the original problem. Woohoo!

To recap, we used the Remainder Theorem to set up a system of equations, solved for the coefficients of the remainder polynomial, and arrived at our final answer. This whole process showcases the power of the Remainder Theorem in tackling polynomial problems. In the next section, we'll summarize our steps and highlight some key takeaways.

Key Takeaways and Summary

Okay, guys, let's take a step back and review what we've done. We started with a polynomial problem that seemed a bit daunting, but we broke it down into manageable steps using the Remainder Theorem. This is a classic technique in algebra, and understanding it can help you tackle a variety of similar problems.

Here’s a quick rundown of the steps we took:

  1. Understanding the Problem: We identified the given information, F(-2) = -5 and F(2) = 3, and recognized that the remainder when dividing by x² - 4 would be in the form ax + b.
  2. Applying the Remainder Theorem: We expressed the division as F(x) = (x² - 4)Q(x) + ax + b and plugged in x = -2 and x = 2 to create a system of equations.
  3. Solving for the Remainder: We solved the system of equations to find a = 2 and b = -1, giving us the remainder R(x) = 2x - 1.

The key takeaway here is the power of the Remainder Theorem. It allows us to connect the remainders of polynomial divisions to the coefficients of the remainder polynomial. This connection is what makes it possible to solve these types of problems systematically.

Another important point is the idea of breaking down a complex problem into smaller, more manageable steps. We didn't try to solve the whole thing at once. Instead, we focused on understanding the givens, representing the remainder correctly, and then using the Remainder Theorem to create equations. This step-by-step approach is a valuable strategy in mathematics and in problem-solving in general.

Finally, practice makes perfect! The more you work through problems like this, the more comfortable you'll become with the techniques and the underlying concepts. So, keep practicing, keep asking questions, and keep exploring the world of polynomials! And remember, the remainder theorem is your friend when it comes to these kinds of problems.