Solving For X³ - 3x² - 5x + 3 Given X = 1/(√5 - 2)

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In the realm of mathematics, we often encounter intricate expressions that require a systematic approach to unravel. One such problem involves finding the value of a polynomial expression given a specific value for the variable. This article delves into the process of evaluating the expression x³ - 3x² - 5x + 3, where x is defined as 1/(√5 - 2). We'll break down the steps, employing algebraic manipulation and simplification techniques to arrive at the solution. This exploration will not only provide the answer but also highlight the importance of strategic problem-solving in mathematics.

Simplifying the Value of x

Before we can substitute the value of x into the polynomial, we need to simplify it first. The given value of x is 1/(√5 - 2). The presence of a radical in the denominator suggests that we should rationalize it. Rationalizing the denominator involves eliminating the radical from the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. In this case, the conjugate of (√5 - 2) is (√5 + 2). Let's perform this operation:

x = 1/(√5 - 2) * (√5 + 2)/(√5 + 2)

Multiplying the numerators, we get 1 * (√5 + 2) = √5 + 2. Multiplying the denominators, we use the difference of squares formula, (a - b)(a + b) = a² - b², where a = √5 and b = 2. Thus, the denominator becomes (√5)² - 2² = 5 - 4 = 1. Therefore, the simplified value of x is:

x = √5 + 2

This simplified form of x will be much easier to work with when we substitute it into the polynomial expression.

Evaluating the Polynomial Expression

Now that we have a simplified value for x, which is √5 + 2, we can substitute it into the polynomial expression x³ - 3x² - 5x + 3. This involves replacing every instance of x in the expression with (√5 + 2). The process may seem daunting at first, but we'll tackle it step by step.

First, let's calculate x²:

x² = (√5 + 2)²

Expanding this using the formula (a + b)² = a² + 2ab + b², we get:

x² = (√5)² + 2(√5)(2) + 2²

x² = 5 + 4√5 + 4

x² = 9 + 4√5

Next, we need to calculate x³:

x³ = x² * x

x³ = (9 + 4√5)(√5 + 2)

Expanding this product, we get:

x³ = 9(√5) + 9(2) + 4√5(√5) + 4√5(2)

x³ = 9√5 + 18 + 4(5) + 8√5

x³ = 9√5 + 18 + 20 + 8√5

x³ = 17√5 + 38

Now we have the values for x, x², and x³. We can substitute these into the original polynomial expression:

x³ - 3x² - 5x + 3 = (17√5 + 38) - 3(9 + 4√5) - 5(√5 + 2) + 3

Distribute the constants:

= 17√5 + 38 - 27 - 12√5 - 5√5 - 10 + 3

Combine like terms:

= (17√5 - 12√5 - 5√5) + (38 - 27 - 10 + 3)

= (17 - 12 - 5)√5 + (38 - 27 - 10 + 3)

= 0√5 + 4

= 4

Therefore, the value of the expression x³ - 3x² - 5x + 3 when x = 1/(√5 - 2) is 4.

Alternative Method: Polynomial Long Division

Another approach to solving this problem involves polynomial long division. This method leverages the fact that if x = √5 + 2, then x - 2 = √5. Squaring both sides, we get (x - 2)² = 5, which expands to x² - 4x + 4 = 5. Rearranging, we have x² - 4x - 1 = 0. This gives us a quadratic equation that x satisfies.

Now, we can perform polynomial long division, dividing the expression x³ - 3x² - 5x + 3 by x² - 4x - 1. This will give us a quotient and a remainder. The remainder will be a linear expression in x, which we can then evaluate using x = √5 + 2. Let's perform the long division:

 x + 1
 x² - 4x - 1 | x³ - 3x² - 5x + 3
 - (x³ - 4x² - x)
 ------------------
 x² - 4x + 3
 - (x² - 4x - 1)
 ------------------
 4

The quotient is x + 1, and the remainder is 4. Therefore, we can write:

x³ - 3x² - 5x + 3 = (x² - 4x - 1)(x + 1) + 4

Since we know that x² - 4x - 1 = 0, the expression simplifies to:

x³ - 3x² - 5x + 3 = (0)(x + 1) + 4

x³ - 3x² - 5x + 3 = 4

This alternative method confirms our previous result, demonstrating the power and versatility of algebraic techniques. Polynomial long division provides a structured way to simplify expressions and can be particularly useful when dealing with higher-degree polynomials.

Key Concepts and Takeaways

This problem illustrates several key concepts in mathematics, including:

  • Rationalizing the denominator: Eliminating radicals from the denominator of a fraction.
  • Algebraic manipulation: Using algebraic rules and identities to simplify expressions.
  • Substitution: Replacing variables with their values to evaluate expressions.
  • Polynomial long division: A method for dividing polynomials.

The solution highlights the importance of simplifying expressions before performing complex calculations. By rationalizing the denominator and simplifying x, we made the substitution process much easier. Furthermore, the alternative method using polynomial long division demonstrates a different approach to the problem, reinforcing the idea that there are often multiple ways to solve a mathematical problem.

In conclusion, by systematically applying algebraic techniques and problem-solving strategies, we successfully evaluated the expression x³ - 3x² - 5x + 3 when x = 1/(√5 - 2). The result, 4, underscores the beauty and precision of mathematical reasoning. This exercise not only provides a solution but also enhances our understanding of fundamental algebraic concepts and techniques.

Applications and Extensions

This type of problem has applications in various areas of mathematics, including calculus, where polynomial functions are frequently encountered. Understanding how to manipulate and evaluate polynomial expressions is crucial for solving a wide range of mathematical problems.

Furthermore, this problem can be extended in several ways. For example, we could explore the roots of the polynomial x³ - 3x² - 5x + 3 and their relationship to the given value of x. We could also investigate the behavior of the polynomial function as x varies, looking for maximum and minimum values. These extensions would provide a deeper understanding of the concepts involved and further develop problem-solving skills.

Conclusion

In this article, we embarked on a journey to evaluate the polynomial expression x³ - 3x² - 5x + 3, given the value of x as 1/(√5 - 2). Through careful algebraic manipulation, simplification, and substitution, we arrived at the solution, 4. We also explored an alternative method using polynomial long division, which reinforced our understanding of the problem and provided a different perspective. This exercise highlights the importance of systematic problem-solving and the power of algebraic techniques in mathematics. The concepts and skills developed in this problem are applicable in various areas of mathematics and can be extended to more complex problems.