Riley's Mistake In Homework Step 2 A Polynomial Fraction Error Analysis

In the realm of mathematics, precision is paramount. Even a seemingly minor misstep can cascade into a significant error, altering the entire outcome of a problem. In this article, we delve into a specific scenario where a student, Riley, encounters a mistake while working on her homework. The problem at hand involves simplifying a complex expression comprising polynomial fractions. Our mission is to meticulously dissect each step of Riley's solution, pinpoint the exact location of the error, and elucidate the underlying mathematical principles that govern the process.

The problem Riley is tackling is:

$\frac{x}{x^2-5 x+6}+\frac{3}{x+3}$

This expression presents a classic challenge in algebraic manipulation: combining two fractions with polynomial denominators. The key to success lies in identifying a common denominator, which allows us to merge the fractions into a single, simplified expression. Let's embark on a step-by-step journey through Riley's attempt, carefully scrutinizing each move to unearth the elusive error.

The initial step in simplifying this expression involves factoring the quadratic denominator of the first fraction. This is a crucial move because it allows us to identify the common factors that will form the basis of our common denominator. Factoring the quadratic expression $x^2 - 5x + 6$ requires us to find two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Therefore, the quadratic expression can be factored as $(x - 2)(x - 3)$.

Riley correctly executes this step, transforming the original expression into:

$\frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$

This transformation sets the stage for finding a common denominator, which is the next critical step in simplifying the expression. The factored form of the denominator provides a clearer picture of the factors that need to be included in the common denominator.

Step 2 is where the pivotal error occurs. Riley's attempt to find a common denominator appears to go astray, leading to an incorrect expression. Let's carefully examine her work:

Step 2: $\frac{x}{(x-2)(x+3)}+\frac{3(x-2)}{(x-2)(x+3)}$

At first glance, this step might seem reasonable. Riley appears to be attempting to create a common denominator by multiplying the second fraction by a suitable factor. However, a closer inspection reveals a critical flaw. The original expression from Step 1 is:

$\frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$

The denominator of the first fraction is $(x - 2)(x - 3)$, while the denominator of the second fraction is $(x + 3)$. To find a common denominator, we need to identify all the unique factors present in both denominators. These factors are $(x - 2)$, $(x - 3)$, and $(x + 3)$. Therefore, the least common denominator (LCD) should be the product of these factors: $(x - 2)(x - 3)(x + 3)$.

However, in Step 2, Riley incorrectly replaces $(x-3)$ with $(x+3)$ in the denominator of the first fraction. This is a fundamental error in algebraic manipulation. The correct approach would have been to multiply the numerator and denominator of the first fraction by $(x + 3)$ and the numerator and denominator of the second fraction by $(x - 2)(x - 3)$.

The correct transformation of the expression should have been:

$\frac{x(x+3)}{(x-2)(x-3)(x+3)}+\frac{3(x-2)(x-3)}{(x+3)(x-2)(x-3)}$

This would have resulted in a common denominator of $(x - 2)(x - 3)(x + 3)$, allowing for the fractions to be combined correctly.

Therefore, Riley's mistake in Step 2 was incorrectly changing the factor $(x-3)$ to $(x+3)$ in the denominator of the first fraction. This error fundamentally alters the expression and leads to an incorrect simplification.

Step 3, which we won't explicitly show here, would inevitably be incorrect due to the error introduced in Step 2. The incorrect common denominator would lead to an incorrect combination of the numerators, resulting in a flawed final expression. This highlights the importance of accuracy in each step of algebraic manipulation; a single error can propagate through the entire solution.

To solidify our understanding and provide a clear contrast to Riley's error, let's walk through the correct solution step-by-step.

Step 1: As before, the first step is to factor the quadratic denominator:

$\frac{x}{x^2-5 x+6}+\frac{3}{x+3} = \frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$

Step 2: Now, we need to find the least common denominator (LCD). As discussed earlier, the LCD is $(x - 2)(x - 3)(x + 3)$. To achieve this common denominator, we multiply the numerator and denominator of each fraction by the missing factors:

$\frac{x}{(x-2)(x-3)} * \frac{(x+3)}{(x+3)} + \frac{3}{(x+3)} * \frac{(x-2)(x-3)}{(x-2)(x-3)}$

This gives us:

$\frac{x(x+3)}{(x-2)(x-3)(x+3)}+\frac{3(x-2)(x-3)}{(x+3)(x-2)(x-3)}$

Step 3: Now that we have a common denominator, we can combine the numerators:

$\frac{x(x+3) + 3(x-2)(x-3)}{(x-2)(x-3)(x+3)}$

Step 4: Expand the terms in the numerator:

$\frac{x^2 + 3x + 3(x^2 - 5x + 6)}{(x-2)(x-3)(x+3)}$

$\frac{x^2 + 3x + 3x^2 - 15x + 18}{(x-2)(x-3)(x+3)}$

Step 5: Combine like terms in the numerator:

$\frac{4x^2 - 12x + 18}{(x-2)(x-3)(x+3)}$

Step 6: Factor out a common factor of 2 from the numerator:

$\frac{2(2x^2 - 6x + 9)}{(x-2)(x-3)(x+3)}$

This is the simplified form of the expression. We cannot factor the quadratic expression in the numerator further, so this is our final answer.

Riley's mistake serves as a valuable learning opportunity, highlighting several key principles in algebraic manipulation:

• Accuracy in Factoring: Factoring polynomials correctly is the cornerstone of simplifying rational expressions. A mistake in factoring can derail the entire solution.
• Finding the Correct Common Denominator: Identifying the least common denominator (LCD) is crucial for combining fractions. The LCD must include all unique factors from the denominators of the fractions being combined. The common denominator must be accurately obtained by multiplying each numerator and denominator by the missing factors, ensuring the values of the fractions remain unchanged. Incorrectly altering factors, as Riley did, leads to a flawed solution.
• Attention to Detail: Even a seemingly small error, like a sign change or incorrect multiplication, can have significant consequences. Therefore, meticulous attention to detail is paramount in every step of the process.
• Double-Checking Your Work: It's always wise to double-check each step of your solution to catch any potential errors. This practice can save time and prevent frustration in the long run.

By understanding the nature of Riley's mistake and working through the correct solution, we can reinforce our understanding of algebraic manipulation and develop the skills necessary to tackle similar problems with confidence. Remember, mathematics is a discipline that rewards precision and careful attention to detail. Embrace these qualities, and you'll be well on your way to mastering the art of problem-solving.

Mathematics, at its core, is a journey of exploration and discovery. Every problem presents an opportunity to learn, grow, and refine our understanding of the world around us. While mistakes are inevitable, they are also invaluable learning experiences. By analyzing our errors, understanding their root causes, and diligently working towards correct solutions, we can unlock the beauty and power of mathematical thinking. So, let's embrace the challenges, celebrate the triumphs, and continue to explore the fascinating world of mathematics with curiosity and perseverance.