Problem 91 - College Algebra with Modeling and Visualization

Question

Simplify. $$ \frac{x}{(x-1)^{2}}-\frac{1}{(x-1)(x+3)} $$

Step-by-Step Solution

Answer

\( \frac{(x+1)^2}{(x-1)^2(x+3)} \) is the simplified form.

Step 1: Identify the denominators

The two fractions given are \( \frac{x}{(x-1)^{2}} \) and \( -\frac{1}{(x-1)(x+3)} \). The denominators are \((x-1)^{2}\) and \((x-1)(x+3)\). To combine these fractions, we need a common denominator.

Step 2: Find the least common denominator (LCD)

The least common denominator for \((x-1)^2\) and \((x-1)(x+3)\) is \((x-1)^2(x+3)\), as it is the smallest expression that can accommodate both original denominators.

Step 3: Rewrite each fraction with the LCD

To rewrite the first fraction, multiply numerator and denominator by \((x+3)\): \[ \frac{x(x+3)}{(x-1)^{2}(x+3)} \] For the second fraction, multiply numerator and denominator by \((x-1)\): \[ \frac{1(x-1)}{(x-1)^{2}(x+3)} \]

Step 4: Combine the fractions

Align the new fractions under the common denominator \((x-1)^2(x+3)\):\[ \frac{x(x+3) - (x-1)}{(x-1)^2(x+3)} \]

Step 5: Simplify the numerator

Expand and combine terms in the numerator: \[ x(x+3) - (x-1) = x^2 + 3x - x + 1 = x^2 + 2x + 1 \]

Step 6: Check for further simplification

The numerator \(x^2 + 2x + 1\) is equivalent to \((x+1)^2\). However, since it shares no common factors with the denominator \((x-1)^2(x+3)\), the expression is in its simplest form.

Key Concepts

Understanding Least Common Denominator (LCD)

The concept of the Least Common Denominator (LCD) is essential when dealing with fractions, especially algebraic fractions. The LCD is the smallest expression that can serve as a common base for fractions, allowing us to perform addition or subtraction. Consider the denominators from our exercise:

• \((x-1)^2\)
• \((x-1)(x+3)\)

To find the LCD, we look for the smallest polynomial expression that includes all factors from both denominators. This process involves:

• Identifying the highest powers of each distinct factor present in any denominator
• Combining these into the least complex form that includes all the original denominators

Thus, for the example given, the LCD is \((x-1)^2(x+3)\), as it incorporates all factors from both original denominators. By having a unified base, we can rewrite and combine fractions more easily.

Simplifying Algebraic Fractions

Simplifying algebraic fractions is akin to simplifying regular fractions, but it involves expressions with variables, often leading to more complex operations. Here's how you simplify them:

• Identify any common factors in the numerator and the denominator.
• Cancel out these common factors to reduce the fraction.

In the exercise, after combining the fractions to get a single expression \[ \frac{x^2 + 2x + 1}{(x-1)^2(x+3)} \], we inspected the numerator \(x^2 + 2x + 1\). This simplified further to \((x+1)^2\), but since \((x+1)\) was not present in the denominator, no further simplification was possible. Simplification is complete when there are no common factors left in both the numerator and denominator.

Exploring Polynomial Expressions

Polynomial expressions form the backbone of algebraic manipulation, often appearing in denominators and numerators of fractions. These expressions are composed of variables raised to whole-number exponents and coefficients. In our exercise:

• \((x-1)^2\) and \((x-1)(x+3)\) are polynomial expressions found in denominators.
• \(x^2 + 2x + 1\) is a polynomial expression derived by simplifying the numerator.

To work effectively with these expressions:

• Understand how to factor polynomials, recognizing identities like the square of a binomial.
• Use these factorizations to combine and simplify expressions effectively.

Mastery of polynomials enables you to manage complex algebraic fractions, making operations simpler and reducing the risk of error.