Problem 16 - Elementary Algebra

Question

Add or Subtract the following rational expressions. $$ \frac{6}{b^{2}+6 b+9}-\frac{2}{b^{2}+4 b+4} $$

Step-by-Step Solution

Answer

Answer: The simplified form of the rational expression is $\frac{4b^2+12b+6}{(b+3)^2(b+2)^2}$.

Step 1: Factoring the denominators

To find the LCD, it helps to first factor the denominators of the given fractions: $$ b^2+6b+9=(b+3)(b+3)=(b+3)^2 \\ b^2+4b+4=(b+2)(b+2)=(b+2)^2 $$

Step 2: Finding the LCD

Analyze both denominators and find their least common denominator: $$ \operatorname{LCD}((b+3)^2,(b+2)^2)=(b+3)^2(b+2)^2 $$

Step 3: Creating equivalent fractions

Now, we find equivalent fractions for both terms with the LCD as the new denominator, by multiplying the numerators and denominators by the appropriate factors. For the first term, we need to multiply both the numerator and denominator by (b+2)^2, and for the second term, we need to multiply by (b+3)^2: $$ \frac{6}{(b+3)^2}\cdot\frac{(b+2)^2}{(b+2)^2}-\frac{2}{(b+2)^2}\cdot\frac{(b+3)^2}{(b+3)^2} = \frac{6(b+2)^2}{(b+3)^2(b+2)^2}-\frac{2(b+3)^2}{(b+3)^2(b+2)^2} $$

Step 4: Combining the fractions

Now that both fractions have the same denominator, we can combine the numerators: $$ \frac{6(b+2)^2-2(b+3)^2}{(b+3)^2(b+2)^2} $$

Step 5: Simplifying the result

Expanding the numerators, we get: $$ \frac{6(b^2+4b+4)-2(b^2+6b+9)}{(b+3)^2(b+2)^2} $$ Now, continue simplifying: $$ \frac{6b^2+24b+24-2b^2-12b-18}{(b+3)^2(b+2)^2} $$ Combining like terms, we get: $$ \frac{4b^2+12b+6}{(b+3)^2(b+2)^2} $$ Since the numerator and denominator have no common factors, the expression cannot be simplified further. The final result is: $$ \frac{4b^2+12b+6}{(b+3)^2(b+2)^2} $$

Key Concepts

Factoring Polynomials

Understanding how to factor polynomials is a critical skill in algebra that allows you to simplify expressions and solve equations more easily. In this exercise, we are given two quadratic expressions in the denominators:

• \(b^2+6b+9\)
• \(b^2+4b+4\)

Both polynomials are perfect squares, which means they can be factored into binomials that are squared. This makes our work much simpler.Factoring \(b^2+6b+9\), we find that it equals \((b+3)(b+3)\), or \((b+3)^2\). Similarly, factoring \(b^2+4b+4\) gives us \((b+2)(b+2)\), or \((b+2)^2\). Factoring these expressions helps us understand the structure of the equation better and paves the way for finding the least common denominator (LCD) when working with rational expressions.

Least Common Denominator

The least common denominator (LCD) is crucial when dealing with the addition or subtraction of fractions because it allows fractions to have a common denominator, facilitating their combination.After factoring the denominators, we observe that:

• \((b+3)^2\) is the denominator for the first expression.
• \((b+2)^2\) is the denominator for the second expression.

To find the LCD, we take each unique factor to its highest power present in any denominator. Thus, the LCD for \((b+3)^2\) and \((b+2)^2\) is \((b+3)^2(b+2)^2\). This common denominator allows us to express both fractions in terms of the same base, which is a crucial step before proceeding to combine the fractions.

Combining Fractions

Combining fractions with the same denominator is straightforward—you simply add or subtract the numerators while keeping the denominator the same. However, getting to this point involves several steps if the original denominators are different.From our prior steps, we established an LCD of \((b+3)^2(b+2)^2\). Therefore, we modify both fractions to have this common denominator. For the first term \(\frac{6}{(b+3)^2}\), we multiply both the numerator and the denominator by \((b+2)^2\). For the second term \(\frac{2}{(b+2)^2}\), we multiply both the numerator and the denominator by \((b+3)^2\):

• \(\frac{6(b+2)^2}{(b+3)^2(b+2)^2}\)
• \(-\frac{2(b+3)^2}{(b+3)^2(b+2)^2}\)

With the fractions now having a common denominator, they can be combined into a single rational expression: \(\frac{6(b+2)^2-2(b+3)^2}{(b+3)^2(b+2)^2}\).

Simplifying Expressions

Simplifying expressions is the process of making them as straightforward as possible by combining like terms and reducing fractions if possible.For the expression \(\frac{6(b+2)^2-2(b+3)^2}{(b+3)^2(b+2)^2}\), we expand the numerators to simplify:

• Expand \(6(b^2+4b+4)\) to get \(6b^2+24b+24\).
• Expand \(-2(b^2+6b+9)\) to get \(-2b^2-12b-18\).

We now combine the like terms from these expansions:

• Combine \(6b^2 - 2b^2 = 4b^2\).
• Combine \(24b - 12b = 12b\).
• Combine \(24 - 18 = 6\).

Thus, the simplified form of the expression is \(\frac{4b^2+12b+6}{(b+3)^2(b+2)^2}\). Checking the numerator and denominator shows no common factors, confirming that no further simplification is possible. The expression is now in its simplest form, ready to be interpreted or applied as needed.