Problem 6 - College Algebra with Corequisite Support

Question

For the following exercises, simplify the rational expressions. \(\frac{6 a^{2}-24 a+24}{6 a^{2}-24}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{a-2}{a+2}\).

Step 1: Identify and Simplify the Greatest Common Factor in the Numerator

In the numerator, we have the expression \(6a^2 - 24a + 24\). Identify the greatest common factor, which is 6. Factor this out from each term: \[ 6(a^2 - 4a + 4) \]

Step 2: Simplify the Upper Expression

Now we simplify the expression inside the parentheses: \(a^2 - 4a + 4\). Notice this is a perfect square trinomial, so it can be rewritten as: \[ (a - 2)^2 \] Thus, the numerator becomes: \[ 6(a - 2)^2 \]

Step 3: Factor the Denominator

In the denominator, we have \(6a^2 - 24\). Identify the greatest common factor, which is also 6. Factor this out: \[ 6(a^2 - 4) \]

Step 4: Simplify the Expression in the Denominator

We should further simplify \(a^2 - 4\) because it is a difference of squares. So it can be factored using the identity \(a^2 - b^2 = (a - b)(a + b)\): \[ (a - 2)(a + 2) \]Thus, the denominator becomes: \[ 6(a - 2)(a + 2) \]

Step 5: Cancel Common Factors

The entire expression now looks like this: \[ \frac{6(a - 2)^2}{6(a - 2)(a + 2)} \] Since 6 is in both the numerator and the denominator, you can cancel it out. Additionally, one \((a - 2)\) term can be canceled from both, leaving: \[ \frac{a - 2}{a + 2} \]

Step 6: Present the Simplified Expression

After canceling the common parts, the simplified rational expression is: \[ \frac{a - 2}{a + 2} \] Make sure no further reductions can be made. The expression is now fully simplified.

Key Concepts

Greatest Common Factor

The greatest common factor (GCF) is the largest number that can divide each term in a given expression without leaving any remainder. It's essential in simplifying rational expressions, as it helps to reduce expressions to their simplest forms. In the problem at hand, both the numerator and the denominator had common factors, simplifying the task greatly.Finding the GCF involves several steps:

• Identify the coefficients in the terms of the expression.
• Look for the largest number that can divide these coefficients evenly.
• Factor out the GCF from the expression.

In the numerator of our original problem, the GCF was 6. By factoring 6 from each term, the expression became easier to simplify: \[6(a^2 - 4a + 4)\].Similarly, in the denominator, 6 was also the GCF, leading to:\[6(a^2 - 4)\].Factoring out the GCF is a crucial step to reduce expressions and manage more complex operations in algebra.

Perfect Square Trinomial

A perfect square trinomial is a special form of quadratic expression. It looks like \(a^2 - 2ab + b^2\) or \(a^2 + 2ab + b^2\), which can be factored into \((a - b)^2\) or \((a + b)^2\) respectively. Recognizing this pattern allows you to simplify quadratic expressions more quickly.In the expression given in the original exercise, after factoring out the greatest common factor, the numerator becomes:\(a^2 - 4a + 4\).This is a perfect square trinomial, because it fits the form \((a - b)^2\) with:

• \(a = a\)
• \(b = 2\)

We see that \(a^2 - 4a + 4\) can be rewritten as:\[(a - 2)^2\].Recognizing and rewriting perfect square trinomials help in reducing and solving complex algebraic expressions efficiently.

Difference of Squares

The difference of squares is a pattern where an expression in the form of \(a^2 - b^2\) can be factored into \((a - b)(a + b)\). This identity is extremely useful for simplifying expressions where one term is subtracted from another, and both are perfect squares.In the problem at hand, after factoring out the greatest common factor from the denominator, we have the expression:\(a^2 - 4\).Recognizing this as a difference of squares helps us apply the difference of squares identity. Here, \(a^2\) is the square of \(a\) and \(4\) is the square of \(2\), leading to:

• \((a - 2)(a + 2)\)

This transformation simplifies the problem significantly, revealing common factors that can be canceled, allowing us to reduce the expression further. Understanding and using the difference of squares pattern is vital for efficient problem-solving in algebra.