For The Polynomial W(x) = 3x - (2x^2 - 1)(5 - X), What Is The Value Of W(5) - W(0)?
In the realm of mathematics, polynomials play a fundamental role. Understanding how to evaluate and manipulate them is crucial for various applications. This article will delve into a specific problem involving a polynomial, providing a detailed, step-by-step solution to enhance your comprehension. We'll focus on calculating the difference between the polynomial's values at two distinct points, W(5) and W(0), for the polynomial W(x) = 3x - (2x^2 - 1)(5 - x). This exercise will not only demonstrate the practical application of polynomial evaluation but also reinforce key concepts in algebraic manipulation. By meticulously working through each step, you'll gain a deeper appreciation for the structure and behavior of polynomials.
Problem Statement
Given the polynomial W(x) = 3x - (2x^2 - 1)(5 - x), determine the difference between W(5) and W(0). This problem requires us to first evaluate the polynomial at x = 5 and x = 0, and then subtract the latter value from the former. This process involves substituting the given values into the polynomial expression and simplifying the result. Understanding how to do this is a core skill in algebra and forms the basis for more complex polynomial operations.
Step 1: Expanding the Polynomial
Before evaluating the polynomial, it's often beneficial to expand and simplify the expression. This makes the substitution process easier and reduces the risk of errors. In our case, we need to expand the term (2x^2 - 1)(5 - x). This involves multiplying each term in the first parenthesis by each term in the second parenthesis. Careful attention to the signs is crucial during this step. The expansion allows us to rewrite the polynomial in a more standard form, where terms are arranged in descending order of their exponents. This standard form is not only easier to work with but also provides a clearer picture of the polynomial's degree and leading coefficient, which are important characteristics for further analysis.
Let's start by expanding the product:
(2x^2 - 1)(5 - x) = 2x^2 * 5 + 2x^2 * (-x) + (-1) * 5 + (-1) * (-x)
Simplifying this expression, we get:
= 10x^2 - 2x^3 - 5 + x
Now, substitute this back into the original polynomial:
W(x) = 3x - (10x^2 - 2x^3 - 5 + x)
Distribute the negative sign:
W(x) = 3x - 10x^2 + 2x^3 + 5 - x
Finally, combine like terms and rearrange in descending order of exponents:
W(x) = 2x^3 - 10x^2 + 2x + 5
This expanded form of the polynomial will make the subsequent evaluation steps much simpler and less prone to errors. By organizing the polynomial in this manner, we've essentially prepared it for efficient computation of its values at specific points.
Step 2: Evaluating W(5)
Now that we have the expanded form of the polynomial, W(x) = 2x^3 - 10x^2 + 2x + 5, we can evaluate it at x = 5. This involves substituting 5 for x in the expression and performing the arithmetic operations. It's crucial to follow the order of operations (PEMDAS/BODMAS) to ensure the correct result. Exponents should be calculated before multiplication and addition. Carefully tracking each step will minimize the chance of making a mistake.
Substitute x = 5 into the polynomial:
W(5) = 2(5)^3 - 10(5)^2 + 2(5) + 5
Calculate the powers:
W(5) = 2(125) - 10(25) + 2(5) + 5
Perform the multiplications:
W(5) = 250 - 250 + 10 + 5
Finally, perform the additions and subtractions:
W(5) = 15
Therefore, the value of the polynomial W(x) at x = 5 is 15. This calculation demonstrates the direct application of polynomial evaluation, a fundamental skill in algebra and calculus.
Step 3: Evaluating W(0)
Next, we need to evaluate the polynomial W(x) = 2x^3 - 10x^2 + 2x + 5 at x = 0. This evaluation is often simpler than evaluating at other values because any term multiplied by zero becomes zero. This significantly reduces the computational effort required. Understanding this property can save time and effort when dealing with polynomials, especially in more complex scenarios. The constant term in the polynomial will be the only term that remains when x = 0.
Substitute x = 0 into the polynomial:
W(0) = 2(0)^3 - 10(0)^2 + 2(0) + 5
Calculate the powers and multiplications:
W(0) = 2(0) - 10(0) + 2(0) + 5
W(0) = 0 - 0 + 0 + 5
Finally, perform the addition:
W(0) = 5
Therefore, the value of the polynomial W(x) at x = 0 is 5. This simple evaluation highlights the importance of understanding the behavior of polynomials at specific points, particularly at x = 0, which often reveals the constant term of the polynomial.
Step 4: Calculating the Difference
Now that we have the values of W(5) and W(0), we can calculate the difference W(5) - W(0). This step is straightforward and involves a simple subtraction. However, it's crucial to ensure that the values are subtracted in the correct order, as subtraction is not commutative. The result of this subtraction will give us the final answer to the problem, representing the change in the polynomial's value between the two points.
We found that W(5) = 15 and W(0) = 5. Therefore,
W(5) - W(0) = 15 - 5
Perform the subtraction:
W(5) - W(0) = 10
Thus, the difference between W(5) and W(0) is 10. This final calculation provides the solution to the original problem, demonstrating the entire process of evaluating a polynomial at specific points and finding the difference between those values.
Final Answer
The difference W(5) - W(0) for the polynomial W(x) = 3x - (2x^2 - 1)(5 - x) is 10. This result matches option b) in the problem statement. This comprehensive solution demonstrates the step-by-step process of evaluating polynomials and finding the difference between their values at different points. This skill is fundamental in algebra and has wide-ranging applications in mathematics and other fields. By understanding the process outlined in this article, you can confidently tackle similar problems and gain a deeper appreciation for the behavior of polynomials.
Conclusion
This article has provided a detailed solution to the problem of finding the difference between W(5) and W(0) for the polynomial W(x) = 3x - (2x^2 - 1)(5 - x). We walked through the process of expanding the polynomial, evaluating it at x = 5 and x = 0, and finally calculating the difference. This step-by-step approach not only provides the correct answer but also reinforces the fundamental concepts of polynomial manipulation and evaluation. By mastering these skills, you'll be well-equipped to tackle more complex algebraic problems and gain a deeper understanding of mathematical concepts. The ability to work with polynomials is a valuable asset in various fields, from engineering and physics to computer science and economics. This problem serves as a great example of how algebraic techniques can be applied to solve practical questions and provides a solid foundation for further exploration of mathematical concepts.