Problem 21 - Elementary Algebra

Question

For exercises 13-24, rewrite each expression as an equivalent expression with the given denominator. $$ \frac{2 p}{p^{2}-36} ;(p+6)(p-6)(p+1) $$

Step-by-Step Solution

Answer

The equivalent expression is \(\frac{2p^2 + 2p}{(p+6)(p-6)(p+1)}\).

Step 1: Factor the denominator

Factorize the denominator of the given expression. Notice that: \[ p^{2} - 36 = (p + 6)(p - 6) \]Thus, the original expression is: \[ \frac{2p}{(p+6)(p-6)} \]

Step 2: Determine missing factors for the new denominator

The given denominator is \[(p+6)(p-6)(p+1)\].Compare this to the factored form from Step 1 \[(p+6)(p-6)\].The missing factor is \[p + 1\].

Step 3: Adjust the numerator

To rewrite the expression with the new denominator, multiply both the numerator and denominator of the original fraction by the missing factor \[p + 1\].This gives: \[\frac{2p(p+1)}{(p+6)(p-6)(p+1)}\].

Step 4: Simplify if necessary

Check if the numerator can be expanded or simplified. The numerator becomes \[2p(p + 1) = 2p^2 + 2p\].Therefore, the equivalent expression is: \[\frac{2p^2 + 2p}{(p+6)(p-6)(p+1)}\].

Key Concepts

Factoring

Factoring is a fundamental concept in algebra, especially when dealing with rational expressions. It involves breaking down a complex polynomial into simpler polynomial factors. For example, in our given problem, the denominator \( p^{2} - 36 \) is a polynomial. Using factoring, we recognize this as a difference of squares:
\[ p^{2} - 36 = (p + 6)(p - 6) \]
By factoring, we simplify expressions and make it easier to identify common terms in numerators and denominators. This step is crucial before adding, subtracting, or simplifying rational expressions.

Least Common Denominator

The Least Common Denominator (LCD) is the smallest expression that all denominators share in a set of fractions. When working with algebraic fractions, the LCD simplifies the process of adding and subtracting fractions. In our exercise, the initial denominator is \((p + 6)(p - 6)\). The goal is to rewrite it with a new given denominator \((p+6)(p-6)(p+1)\). Here, we see that the given denominator already includes all the factors of the initial denominator, plus an additional factor - \(p + 1\). By identifying the LCD, we ensure that all expressions have a common baseline for further operations.

Algebraic Fractions

Algebraic fractions are fractions where the numerator, denominator, or both contain algebraic expressions. In this case, our original algebraic fraction is:
\[ \frac{2p}{p^2 - 36} = \frac{2p}{(p+6)(p-6)} \]
To rewrite this fraction with a new denominator \((p+6)(p-6)(p+1)\), we need to maintain the equality by adjusting the numerator accordingly. By multiplying both the numerator and the denominator by the missing factor \(p + 1\) to match the given denominator, we obtain:
\[ \frac{2p(p+1)}{(p+6)(p-6)(p+1)} \]
Algebraic fractions follow the same rules as numerical fractions but require careful handling of algebraic expressions.

Polynomial Expressions

Polynomial expressions consist of variables raised to whole number exponents and their coefficients. In the context of our exercise, there are polynomial expressions in both the numerator and the denominator. Initially, the numerator is a simple polynomial, \(2p\), and the denominator is a factored polynomial, \((p+6)(p-6)\).
To adjust the numerator for the new denominator, we multiply it by the missing factor \(p + 1\). This results in:
\[ 2p(p+1) = 2p^2 + 2p \]
Here, we expanded the product into a standard polynomial form. Understanding polynomial expressions and their operations (like factoring and expansion) is essential for handling rational expressions effectively.