Problem 42 - Elementary Algebra

Question

Simplify each algebraic fraction. $$\frac{6 a^{2}-11 a-10}{8 a^{2}-22 a+5}$$

Step-by-Step Solution

Answer

The simplified form of the fraction is \( \frac{3a+2}{4a-1} \) excluding \( a = \frac{5}{2} \).

Step 1: Understand the Problem

The goal is to simplify the algebraic fraction \( \frac{6a^{2}-11a-10}{8a^{2}-22a+5} \). This involves factoring both the numerator and the denominator to reduce the fraction to its simplest form.

Step 2: Factor the Numerator

The numerator of the fraction is \( 6a^2 - 11a - 10 \). We need to find two numbers whose product is \( 6 \times (-10) = -60 \) and which add up to \( -11 \). These numbers are \( -15 \) and \( 4 \). Thus, the numerator factors as: \( 6a^2 - 15a + 4a - 10 = (3a+2)(2a-5) \).

Step 3: Factor the Denominator

The denominator of the fraction is \( 8a^2 - 22a + 5 \). Find two numbers whose product is \( 8 \times 5 = 40 \) and which add up to \( -22 \). These numbers are \( -20 \) and \( -2 \). Thus, the denominator factors as: \( 8a^2 - 20a - 2a + 5 = (4a-1)(2a-5) \).

Step 4: Simplify the Fraction

Now, we have the fraction: \( \frac{(3a+2)(2a-5)}{(4a-1)(2a-5)} \). Since \( (2a-5) \) is a common factor, we can cancel it out, resulting in the simplified fraction \( \frac{3a+2}{4a-1} \).

Step 5: Verify the Simplification

The process involves ensuring that the simplification is valid by considering the restrictions where the original denominator is zero. \( 2a-5 = 0 \) implies \( a = \frac{5}{2} \). Ensure the solution acknowledges this restriction by noting that the simplified fraction is valid for all \( a eq \frac{5}{2} \).

Key Concepts

Factoring Polynomials

Factoring polynomials means breaking down an expression into products of simpler polynomials, making complex equations easier to work with. Here, it's essential to follow a structured approach:

• Identify terms that can be grouped and factored out.
• Use the product-sum method, where you're looking for two numbers that multiply to a specific product of coefficients, while also summing to another number.

In the given exercise, the numerator \(6a^2 - 11a - 10\) uses this method. First, calculate the product of the first and last coefficients: \(6 \times -10 = -60\). We need numbers that multiply to -60 and add to -11. These numbers are -15 and 4. By rearranging terms and factoring by grouping, we obtain \((3a+2)(2a-5)\).
The same concept applies to the denominator \(8a^2 - 22a + 5\), where the product is \(8 \times 5 = 40\) and the sum is -22. Identify -20 and -2 as the numbers, then factor by grouping, leading to \((4a-1)(2a-5)\).
Thus, each part of the polynomial is expressed in simpler terms, ready to simplify the fraction further.

Simplification

Simplification in algebra involves reducing expressions to their simplest form. The aim is to make them as easy to work with as possible, typically by canceling out common terms.
In this example, after factoring both parts of the algebraic fraction, we end up with the fraction:
\[\frac{(3a+2)(2a-5)}{(4a-1)(2a-5)}\]The factor \((2a-5)\) appears in both the numerator and the denominator. When simplifying fractions, any common factor can be canceled out, provided it doesn’t lead to dividing by zero. Always watch out for such constraints as they might affect the validity of the simplification.
After canceling \((2a-5)\), the simplified expression becomes:
\[\frac{3a+2}{4a-1}\]Simplification ensures that the algebraic fraction is expressed in the most concise form, while still being equivalent to the original expression.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Understanding them is a crucial part of algebra, as they appear often in problems involving rates, ratios, and proportions.

• Key aspects include identifying common factors in the numerator and denominator.
• Simplifying by canceling those common factors, while keeping track of any restrictions on the variables.

For the expression \(\frac{6a^2-11a-10}{8a^2-22a+5}\), both parts are polynomials. By factoring, we simplify the expression while ensuring it remains equivalent.
Another critical point is acknowledging restrictions like the exclusion of values making the denominator zero, here \(a eq \frac{5}{2}\) from \(2a-5=0\). Rationalizing these expressions also prepares you to tackle more advanced topics in algebra, transforming what's complex into something manageable and clearly understandable.