Polynomial Remainder Theorem Problem: Find The Value Of 'a'

Hey guys! Let's dive into a fun math problem today that involves polynomials and the Remainder Theorem. We've got two polynomials, $f(x)$ and $g(x)$, and we're going to use the fact that they have the same remainder when divided by $(x+1)$ to figure out something cool. This is a classic algebra problem, and understanding it will seriously boost your skills in polynomial manipulation. So, buckle up and let's get started!

Understanding the Problem

Okay, so here's the deal. We're given two polynomials:

• $f(x) = x^3 - 2x^2 + 2ax - 6$
• g(x) = x^2 - rac{1}{2}ax + 4x - a

The key piece of information here is that when we divide both of these polynomials by $(x+1)$, they leave the same remainder. This is where the Remainder Theorem comes into play. The Remainder Theorem is a powerful tool in polynomial algebra. It basically says that if you divide a polynomial $f(x)$ by $(x - c)$, the remainder is simply $f(c)$. So, to find the remainder when dividing by $(x + 1)$, we need to evaluate the polynomials at $x = -1$. This seemingly simple concept is what unlocks the solution to our problem. Remember, the Remainder Theorem simplifies finding remainders, saving us from long division. Understanding this is crucial for tackling this kind of problem effectively. The beauty of the Remainder Theorem lies in its directness and how it transforms a division problem into a simple substitution.

Think of it like this: imagine you have a complex machine, and the Remainder Theorem is like a special key that opens a hidden compartment, revealing the answer directly. Without this key, you'd have to разбираться with the entire machine, which in this case is polynomial long division. Now, back to our polynomials. We need to apply this theorem to both $f(x)$ and $g(x)$. This involves substituting $x = -1$ into each polynomial and then using the given information that the remainders are equal to set up an equation. This equation will involve the unknown variable 'a', which is what we want to find. By solving this equation, we'll determine the value of 'a' that makes the remainders the same. It’s like solving a puzzle where the pieces are polynomial expressions and the Remainder Theorem is the guide to fitting them together. So, let's move on to the next step, where we actually put this theorem into action and calculate the remainders.

Applying the Remainder Theorem

Now comes the fun part – putting the Remainder Theorem to work! We know that if we divide $f(x)$ by $(x + 1)$, the remainder is $f(-1)$. Similarly, the remainder when dividing $g(x)$ by $(x + 1)$ is $g(-1)$. Let's calculate these remainders:

For $f(x)$:

$f(-1) = (-1)^3 - 2(-1)^2 + 2a(-1) - 6$

$f(-1) = -1 - 2 - 2a - 6$

$f(-1) = -9 - 2a$

So, the remainder when $f(x)$ is divided by $(x + 1)$ is $-9 - 2a$.

Now, let's do the same for $g(x)$:

g(-1) = (-1)^2 - rac{1}{2}a(-1) + 4(-1) - a

g(-1) = 1 + rac{1}{2}a - 4 - a

g(-1) = -3 - rac{1}{2}a

Thus, the remainder when $g(x)$ is divided by $(x + 1)$ is -3 - rac{1}{2}a. Remember, the goal here is to find the value of 'a' that makes these remainders equal. We've now successfully found expressions for both remainders in terms of 'a'. This is a crucial step because it allows us to set up an equation. Setting up the correct equation is like laying the foundation for a building – if it's not solid, the entire structure might crumble. In this case, the equation will connect the two remainders, reflecting the problem's core condition: that they are the same. Think of it as balancing a scale. On one side, we have the remainder from $f(x)$, and on the other side, the remainder from $g(x)$. We need to adjust the value of 'a' until the scale is perfectly balanced, indicating that the remainders are equal. This leads us directly to the next step: setting the two remainder expressions equal to each other and solving for 'a'.

Solving for 'a'

Okay, we've got the remainders for both $f(x)$ and $g(x)$ in terms of 'a'. The problem tells us these remainders are equal. So, let's set them equal to each other and solve for 'a':

-9 - 2a = -3 - rac{1}{2}a

To solve this equation, we need to get all the 'a' terms on one side and the constants on the other. First, let's add $2a$ to both sides:

-9 = -3 + rac{3}{2}a

Now, add $3$ to both sides:

-6 = rac{3}{2}a

Finally, multiply both sides by rac{2}{3} to isolate 'a':

a = -6 * rac{2}{3}

$a = -4$

So, we've found that $a = -4$. This is a major step! We've successfully solved for 'a' using the information given and the Remainder Theorem. It's like cracking a code – we started with a set of conditions, applied a key concept (the Remainder Theorem), and worked through the algebraic steps to reveal the hidden value. The value of 'a' we've found is the linchpin that connects the two polynomials, ensuring they have the same remainder when divided by $(x + 1)$. But before we celebrate too much, it's always a good idea to double-check our answer. We can do this by plugging $a = -4$ back into the original remainder expressions for $f(x)$ and $g(x)$ and verifying that they are indeed equal. This is like testing a lock with the key we've created – does it actually work? If the remainders match, we can be confident in our solution. This verification step adds an extra layer of assurance, transforming our calculated answer into a confirmed solution.

Verifying the Solution

It's always a good idea to double-check our work, especially in math! Let's plug $a = -4$ back into the remainder expressions we found earlier:

For $f(x)$, the remainder was $-9 - 2a$. Substituting $a = -4$:

Remainder = $-9 - 2(-4) = -9 + 8 = -1$

For $g(x)$, the remainder was -3 - rac{1}{2}a. Substituting $a = -4$:

Remainder = -3 - rac{1}{2}(-4) = -3 + 2 = -1

Great! Both remainders are the same (-1), so our solution $a = -4$ is correct. Woohoo! We successfully navigated the problem and found the value of 'a'. This verification step is not just a formality; it’s a critical part of the problem-solving process. It's like proofreading an essay before submitting it – you want to make sure there are no errors before you finalize your answer. By verifying, we're essentially building a strong case for our solution, showing that it's consistent with all the given information and the principles of the Remainder Theorem. Think of it as adding a seal of approval to our answer, boosting our confidence and ensuring accuracy. Now that we've confirmed our solution, we can confidently move on to the final step, which might involve answering specific questions or making conclusions based on our findings. The journey through this problem highlights the power of the Remainder Theorem and the importance of careful algebraic manipulation and verification.

Conclusion

So, we started with two polynomials and the information that they have the same remainder when divided by $(x + 1)$. We used the Remainder Theorem to find expressions for these remainders in terms of 'a', set those expressions equal, solved for 'a', and then verified our solution. The value of $a$ that satisfies the given condition is $a = -4$. Awesome job working through this problem with me! You've now got a solid grasp of how to apply the Remainder Theorem in this type of scenario. Remember, the key is to break down the problem into smaller, manageable steps. First, understand the given information and the relevant theorems (like the Remainder Theorem). Second, apply the theorem to set up equations. Third, solve the equations carefully, paying attention to algebraic manipulations. And finally, always verify your solution to ensure accuracy. This problem-solving approach is not just useful in mathematics; it's a valuable skill that can be applied in various aspects of life. Think of each problem as a challenge, and the tools and techniques you learn as the keys to unlocking the solution. By practicing and persevering, you'll become a more confident and effective problem-solver. Keep up the great work, and remember, math can be fun and rewarding when you approach it with a positive attitude and a willingness to learn!