Determine The Remainder When F ( X ) = 3 X 5 − 5 X 2 + 4 X + 1 F(x) = 3x^5 - 5x^2 + 4x + 1 F ( X ) = 3 X 5 − 5 X 2 + 4 X + 1 Is Divided By ( X − 1 ) ( X + 2 ) (x-1)(x+2) ( X − 1 ) ( X + 2 )

Jun 17, 2025 by ADMIN 190 views

Determining the Remainder of a Polynomial Division

When dealing with polynomial division, a crucial concept is finding the remainder. This article delves into the process of determining the remainder when a polynomial, specifically f(x) = 3x⁵ - 5x² + 4x + 1, is divided by a quadratic divisor, (x - 1)(x + 2). This topic is fundamental in algebra and precalculus, particularly when exploring polynomial functions and the Remainder Theorem. Understanding how to efficiently find remainders is essential for simplifying complex expressions and solving various mathematical problems. We will explore the Remainder Theorem and Polynomial Long Division, showcasing how to apply these methods effectively. Our detailed explanations and step-by-step examples will provide a comprehensive understanding of polynomial division and remainder determination.

Understanding the Remainder Theorem

The Remainder Theorem is a cornerstone of polynomial algebra. It states that when a polynomial p(x) is divided by a linear factor (x - a), the remainder is p(a). This theorem provides a quick and efficient way to find the remainder without performing full polynomial long division. It's a powerful tool for simplifying polynomial expressions and solving related problems. For instance, if we divide a polynomial by (x - 2), we can simply evaluate the polynomial at x = 2 to find the remainder. The Remainder Theorem is not only a shortcut but also a fundamental concept that underpins many polynomial operations. Its applications extend to factorization, root finding, and various other algebraic manipulations. Furthermore, understanding the Remainder Theorem enhances problem-solving skills in advanced mathematics, making it a crucial topic for students and educators alike. This theorem lays the groundwork for more complex polynomial divisions and remainder calculations, bridging the gap between simple linear divisors and more complex quadratic or higher-degree divisors.

To fully grasp the Remainder Theorem, it's essential to understand its connection to the Division Algorithm. The Division Algorithm states that for any polynomials p(x) and d(x) (where d(x) is not zero), there exist unique polynomials q(x) (the quotient) and r(x) (the remainder) such that p(x) = d(x)q(x) + r(x), where the degree of r(x) is less than the degree of d(x). When d(x) is a linear factor (x - a), the remainder r(x) is a constant, which is exactly what the Remainder Theorem helps us find. This constant is the value of the polynomial p(x) evaluated at x = a. By applying the Remainder Theorem, we bypass the need for long division in many cases, streamlining the process of finding remainders. This theorem is particularly useful when dealing with higher-degree polynomials and complex divisions, where manual division can be cumbersome and time-consuming. The Remainder Theorem, therefore, serves as a valuable tool in simplifying polynomial algebra and facilitating efficient problem-solving.

Applying the Remainder Theorem to Linear Factors

When applying the Remainder Theorem, it is critical to accurately identify the value of 'a' from the linear factor (x - a). For example, if the divisor is (x - 1), then a = 1, and if the divisor is (x + 2), then a = -2. This seemingly simple step is foundational, as substituting the wrong value will lead to an incorrect remainder. Once the value of 'a' is correctly identified, the next step is to substitute it into the polynomial f(x). This substitution involves replacing every instance of x in the polynomial with the value of 'a'. The resulting numerical expression is then evaluated using the order of operations (PEMDAS/BODMAS). It's crucial to perform these calculations meticulously, paying close attention to signs and exponents, to ensure an accurate remainder. This method of substitution and evaluation is not only efficient but also reduces the likelihood of errors compared to polynomial long division, especially for higher-degree polynomials. Moreover, understanding this process strengthens the grasp of polynomial functions and their behavior, which is essential for more advanced algebraic concepts.

Consider our given polynomial, f(x) = 3x⁵ - 5x² + 4x + 1. To find the remainder when f(x) is divided by (x - 1), we substitute x = 1 into f(x): f(1) = 3(1)⁵ - 5(1)² + 4(1) + 1 = 3 - 5 + 4 + 1 = 3. Thus, the remainder when f(x) is divided by (x - 1) is 3. Similarly, to find the remainder when f(x) is divided by (x + 2), we substitute x = -2 into f(x): f(-2) = 3(-2)⁵ - 5(-2)² + 4(-2) + 1 = 3(-32) - 5(4) - 8 + 1 = -96 - 20 - 8 + 1 = -123. Therefore, the remainder when f(x) is divided by (x + 2) is -123. These calculations demonstrate the straightforward application of the Remainder Theorem to find remainders with linear divisors. The method is concise and reliable, making it an indispensable tool in polynomial algebra. This approach not only simplifies problem-solving but also enhances comprehension of polynomial behavior and function evaluation.

Addressing Division by a Quadratic Factor

When dividing by a quadratic factor like (x - 1)(x + 2), the Remainder Theorem alone isn't directly applicable in the same way as with linear factors. The Remainder Theorem provides the remainder when dividing by a linear factor, but for quadratic divisors, the remainder can be a linear expression. This is because the degree of the remainder must be less than the degree of the divisor. Since the divisor is quadratic (degree 2), the remainder can be at most linear (degree 1). Therefore, we express the remainder as rx + s, where r and s are constants to be determined. To find these constants, we leverage the remainders obtained when dividing by the linear factors of the quadratic divisor. This involves setting up a system of equations based on the remainders found using the Remainder Theorem for each linear factor. Solving this system of equations allows us to determine the values of r and s, thus defining the remainder when dividing by the quadratic factor. This method provides a systematic approach to handling divisions by higher-degree polynomials, showcasing the interconnectedness of algebraic concepts and techniques.

To elaborate, when we divide f(x) by (x - 1)(x + 2), we express the division using the Division Algorithm: f(x) = (x - 1)(x + 2)q(x) + (rx + s), where q(x) is the quotient and rx + s is the remainder. We already know the remainders when f(x) is divided by (x - 1) and (x + 2) individually. These remainders give us two crucial pieces of information. When x = 1, f(1) = 3, and the equation becomes 3 = r(1) + s, which simplifies to r + s = 3. Similarly, when x = -2, f(-2) = -123, and the equation becomes -123 = r(-2) + s, which simplifies to -2r + s = -123. Now we have a system of two linear equations with two variables, r and s. Solving this system will give us the coefficients of the remainder. This process demonstrates the power of combining different algebraic techniques to solve complex problems. By leveraging the Remainder Theorem and the Division Algorithm, we can efficiently determine the remainder when dividing by a quadratic factor. This method is not only mathematically sound but also provides a clear and structured approach to problem-solving.

Solving for the Remainder

To solve the system of equations, r + s = 3 and -2r + s = -123, we can use various methods such as substitution or elimination. The elimination method is particularly effective here. Subtracting the second equation from the first eliminates s, allowing us to solve for r. This step is crucial as it simplifies the problem into a single-variable equation. Once we find the value of r, we can substitute it back into either of the original equations to find the value of s. This process of solving a system of linear equations is a fundamental skill in algebra and is widely applicable in various mathematical contexts. Accurate algebraic manipulation is key to obtaining the correct values for r and s, which in turn define the remainder when dividing by the quadratic factor. The ability to solve such systems efficiently not only enhances problem-solving speed but also reinforces the understanding of linear algebra principles. Moreover, this approach highlights the interconnectedness of different algebraic concepts and techniques, showcasing how they can be combined to tackle more complex problems.

Subtracting the equation -2r + s = -123 from r + s = 3, we get (r + s) - (-2r + s) = 3 - (-123), which simplifies to 3r = 126. Dividing both sides by 3, we find r = 42. Now, substituting r = 42 into the equation r + s = 3, we get 42 + s = 3. Subtracting 42 from both sides, we find s = -39. Therefore, the remainder is rx + s = 42x - 39. This result indicates that when f(x) = 3x⁵ - 5x² + 4x + 1 is divided by (x - 1)(x + 2), the remainder is the linear expression 42x - 39. This comprehensive solution demonstrates the application of the Remainder Theorem and the solution of a system of linear equations to determine the remainder when dividing by a quadratic factor. The process underscores the importance of algebraic manipulation and problem-solving strategies in polynomial algebra. This approach not only provides the answer but also enhances the understanding of the underlying mathematical principles.

Polynomial Long Division as an Alternative Method

Polynomial long division provides an alternative method for finding the remainder when dividing by a quadratic factor. While it may be more laborious than the method combining the Remainder Theorem and solving a system of equations, it is a fundamental technique in polynomial algebra. This method mirrors the process of long division with numbers, but instead of digits, we work with terms of the polynomials. Polynomial long division involves dividing the dividend (the polynomial being divided) by the divisor (in this case, the quadratic factor) step by step, determining the quotient and the remainder. The process requires careful attention to detail, including aligning terms with the same degree and correctly performing multiplication and subtraction. Despite its complexity, polynomial long division provides a comprehensive understanding of polynomial division and serves as a valuable tool when the Remainder Theorem cannot be directly applied. Furthermore, mastering this technique enhances algebraic proficiency and provides a solid foundation for more advanced mathematical concepts.

To perform polynomial long division, we set up the division problem with the dividend (3x⁵ - 5x² + 4x + 1) and the divisor (x - 1)(x + 2), which expands to x² + x - 2. The process begins by dividing the leading term of the dividend (3x⁵) by the leading term of the divisor (x²), resulting in 3x³. This is the first term of the quotient. We then multiply the divisor by 3x³ and subtract the result from the dividend. This step is crucial for reducing the degree of the polynomial. We continue this process, bringing down the next term and dividing the new leading term by the leading term of the divisor until the degree of the remainder is less than the degree of the divisor. The final non-zero polynomial left after the last subtraction is the remainder. Polynomial long division, although more intricate than using the Remainder Theorem, offers a visual and step-by-step approach to polynomial division. It reinforces the understanding of polynomial structure and manipulation, providing a deeper insight into the division process. This method not only provides the remainder but also the quotient, making it a comprehensive tool for polynomial division.

The steps for polynomial long division are as follows:

Divide 3x⁵ by x² to get 3x³.
Multiply (x² + x - 2) by 3x³ to get 3x⁵ + 3x⁴ - 6x³.
Subtract (3x⁵ + 3x⁴ - 6x³) from (3x⁵ - 5x² + 4x + 1) to get -3x⁴ + 6x³ - 5x² + 4x + 1.
Divide -3x⁴ by x² to get -3x².
Multiply (x² + x - 2) by -3x² to get -3x⁴ - 3x³ + 6x².
Subtract (-3x⁴ - 3x³ + 6x²) from (-3x⁴ + 6x³ - 5x² + 4x + 1) to get 9x³ - 11x² + 4x + 1.
Divide 9x³ by x² to get 9x.
Multiply (x² + x - 2) by 9x to get 9x³ + 9x² - 18x.
Subtract (9x³ + 9x² - 18x) from (9x³ - 11x² + 4x + 1) to get -20x² + 22x + 1.
Divide -20x² by x² to get -20.
Multiply (x² + x - 2) by -20 to get -20x² - 20x + 40.
Subtract (-20x² - 20x + 40) from (-20x² + 22x + 1) to get 42x - 39.

The remainder is 42x - 39, which matches the result obtained using the Remainder Theorem and solving the system of equations. This verification reinforces the accuracy of both methods and highlights the versatility of polynomial algebra techniques. Polynomial long division serves as a robust method for handling polynomial division problems, particularly when dealing with higher-degree polynomials or when a comprehensive view of the division process is desired.

Conclusion

In conclusion, finding the remainder when f(x) = 3x⁵ - 5x² + 4x + 1 is divided by (x - 1)(x + 2) can be achieved through two primary methods: the combination of the Remainder Theorem and solving a system of equations, and polynomial long division. The former method provides an efficient approach by leveraging the remainders obtained from linear factors and setting up a system of equations to determine the coefficients of the remainder. The latter, polynomial long division, offers a more direct but often lengthier process that mirrors numerical long division. Both methods yield the same remainder, 42x - 39, demonstrating the consistency and versatility of algebraic techniques. Understanding these methods is crucial for mastering polynomial algebra and solving a wide range of mathematical problems. Whether one prefers the elegance of the Remainder Theorem or the methodical nature of long division, proficiency in both enhances problem-solving capabilities and deepens the understanding of polynomial functions. This exploration underscores the importance of mastering fundamental algebraic concepts and techniques, which are essential for advanced mathematical studies and applications.