Remainder Theorem: Is C A Zero Of The Polynomial?
Hey guys! Today, we're diving into the Remainder Theorem and how we can use it to figure out if a given complex number c is a zero of a polynomial. We'll be working with the polynomial n(x) = 2x³ - 3x² + 60x + 200, and we're going to check two values for c: (a) c = 4 + 5i and (b) c = 2 + i. Buckle up, because this is going to be a fun ride!
Understanding the Remainder Theorem
Before we jump into the calculations, let's quickly recap what the Remainder Theorem is all about. In essence, the Remainder Theorem states that if you divide a polynomial n(x) by (x - c), the remainder you get is equal to n(c). This is super useful because if n(c) = 0, then c is a zero (or root) of the polynomial. This means that (x - c) is a factor of n(x). So, our mission today is to plug in our values for c into the polynomial and see if we get zero. If we do, bingo, we've found a zero!
Now, why is this important? Well, finding zeros of polynomials is a fundamental problem in algebra. Zeros tell us a lot about the behavior of the polynomial, like where it crosses the x-axis, and they're crucial for solving polynomial equations. The Remainder Theorem provides an elegant and efficient way to check potential zeros, especially when dealing with complex numbers. This theorem isn't just a mathematical curiosity; it's a powerful tool in our problem-solving arsenal. It simplifies the process of finding roots and factors, which are essential in various applications of polynomial functions. From engineering to computer graphics, understanding polynomial roots helps us model and analyze real-world phenomena. The ability to quickly determine if a number is a root allows for faster problem-solving and more efficient use of computational resources.
So, let’s get our hands dirty and see how this theorem works in practice. Remember, our goal is to substitute the given complex numbers into the polynomial and meticulously calculate the result. We'll be dealing with complex arithmetic, so precision is key. A single mistake in our calculations can lead us down the wrong path. We'll break down each step, ensuring we follow the order of operations and handle complex numbers with care. By the end of this process, we'll not only have applied the Remainder Theorem but also sharpened our skills in complex number manipulation, a valuable skill in many areas of mathematics and beyond. So, let's proceed with the confidence that comes from understanding the theory and the diligence to execute the calculations flawlessly.
(a) Checking if c = 4 + 5i is a Zero
Let's start with our first candidate, c = 4 + 5i. We need to find n(4 + 5i). This means we're going to substitute (4 + 5i) for every x in our polynomial: n(x) = 2x³ - 3x² + 60x + 200. Get ready for some complex number arithmetic!
First, let's find (4 + 5i)². Remember that i² = -1, so we'll need to keep that in mind as we expand:
(4 + 5i)² = (4 + 5i)(4 + 5i) = 16 + 20i + 20i + 25i² = 16 + 40i - 25 = -9 + 40i
Next, we need (4 + 5i)³. We can calculate this by multiplying (4 + 5i)² by (4 + 5i):
(4 + 5i)³ = (-9 + 40i)(4 + 5i) = -36 - 45i + 160i + 200i² = -36 + 115i - 200 = -236 + 115i
Now we can plug these values back into our polynomial:
n(4 + 5i) = 2(-236 + 115i) - 3(-9 + 40i) + 60(4 + 5i) + 200
Distribute and simplify:
n(4 + 5i) = -472 + 230i + 27 - 120i + 240 + 300i + 200
Combine the real and imaginary parts:
n(4 + 5i) = (-472 + 27 + 240 + 200) + (230 - 120 + 300)i = -5 + 410i
So, n(4 + 5i) = -5 + 410i. Since this is not equal to zero, c = 4 + 5i is not a zero of the polynomial.
This step-by-step breakdown is crucial when dealing with complex numbers. The potential for errors in arithmetic is high, and a systematic approach helps maintain accuracy. Each calculation, from squaring the complex number to cubing it and then substituting into the polynomial, must be meticulously executed. The distribution and simplification stages are also critical, ensuring that like terms are correctly combined. The final result, a complex number with both real and imaginary parts, must then be carefully evaluated to determine if it equals zero. The fact that the result is not zero is a significant finding, indicating that the initial guess of (4 + 5i) as a root was incorrect. This process underscores the importance of the Remainder Theorem as a tool for efficiently checking potential roots without resorting to more cumbersome methods like polynomial long division.
(b) Checking if c = 2 + i is a Zero
Alright, let's move on to our second candidate, c = 2 + i. We're going to follow the same process as before, but with a different complex number. We need to calculate n(2 + i), so we'll substitute (2 + i) for every x in our polynomial n(x) = 2x³ - 3x² + 60x + 200.
First, let's find (2 + i)²:
(2 + i)² = (2 + i)(2 + i) = 4 + 2i + 2i + i² = 4 + 4i - 1 = 3 + 4i
Now, let's calculate (2 + i)³. We'll multiply (2 + i)² by (2 + i):
(2 + i)³ = (3 + 4i)(2 + i) = 6 + 3i + 8i + 4i² = 6 + 11i - 4 = 2 + 11i
Now we plug these values back into our polynomial:
n(2 + i) = 2(2 + 11i) - 3(3 + 4i) + 60(2 + i) + 200
Distribute and simplify:
n(2 + i) = 4 + 22i - 9 - 12i + 120 + 60i + 200
Combine the real and imaginary parts:
n(2 + i) = (4 - 9 + 120 + 200) + (22 - 12 + 60)i = 315 + 70i
So, n(2 + i) = 315 + 70i. Since this is also not equal to zero, c = 2 + i is not a zero of the polynomial either.
The process for evaluating n(2 + i) mirrors the approach used for n(4 + 5i), but it reinforces the nuances of complex number arithmetic. Squaring (2 + i) and cubing it involve similar expansions and simplifications using the property i² = -1. The critical step here is the meticulous application of the distributive property and the accurate grouping of real and imaginary terms. When substituting these values back into the original polynomial, it is vital to maintain the correct order of operations and signs. The final result, 315 + 70i, once again demonstrates that the value of the polynomial at this complex number is not zero, leading us to conclude that (2 + i) is not a root of the polynomial. This consistent application of the Remainder Theorem with different complex numbers reinforces the importance of accuracy and systematic calculation in complex analysis. Furthermore, it highlights the theorem's effectiveness in quickly ruling out potential roots without engaging in more time-consuming division methods.
Conclusion
So, there you have it! We've used the Remainder Theorem to check if c = 4 + 5i and c = 2 + i are zeros of the polynomial n(x) = 2x³ - 3x² + 60x + 200. We found that neither of them are zeros because n(4 + 5i) = -5 + 410i and n(2 + i) = 315 + 70i, both of which are not equal to zero.
This exercise demonstrates the power and efficiency of the Remainder Theorem in determining whether a given number is a root of a polynomial. It’s a crucial tool in algebra, especially when dealing with complex numbers. Remember, the key is to carefully substitute and simplify, keeping track of those pesky i² = -1 terms. Keep practicing, and you'll become a pro at using the Remainder Theorem!
In summary, the Remainder Theorem provides a straightforward method to evaluate whether a number is a root of a polynomial, particularly valuable when dealing with complex numbers. By substituting the number into the polynomial and calculating the result, we can quickly determine if the remainder is zero, indicating a root. This process avoids the complexities of polynomial division, making it an efficient tool for root identification. Both evaluations in this example highlighted the necessity of meticulous calculations and careful handling of complex arithmetic. The non-zero results obtained in both cases underscore the precision required when working with complex numbers and polynomials. Understanding and applying the Remainder Theorem not only enhances our problem-solving capabilities in algebra but also lays a solid foundation for more advanced mathematical concepts and applications in various fields.