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Magazine Chapter 8: Problem 254
Simplify the expression \(\frac{4}{n^{2}+6 n+9}-\frac{1}{n^{2}-9}\) and explain all your steps.
Short Answer
Expert verified
\(\frac{3(n-5)}{(n+3)^2(n-3)}\)
Step by step solution
01
Factor the Denominators
First, factor the denominators. The denominator of the first fraction is a perfect square trinomial: \[n^2 + 6n + 9 = (n + 3)^2\]For the second fraction, the denominator is a difference of squares: \[n^2 - 9 = (n - 3)(n + 3)\]
02
Rewrite the Fractions with Factored Denominators
Write the fractions with their denominators factored. The expression becomes: \[\frac{4}{(n+3)^2} - \frac{1}{(n-3)(n+3)}\]
03
Find a Common Denominator
Identify the common denominator, which is the least common multiple of \((n+3)^2\)and \((n-3)(n+3)\). The common denominator is \((n+3)^2(n-3)\).
04
Rewrite Each Fraction with the Common Denominator
Rewrite each fraction so they have the common denominator \((n+3)^2(n-3)\). This gives: \[\frac{4(n-3)}{(n+3)^2(n-3)} - \frac{1(n+3)}{(n-3)(n+3)^2} = \frac{4(n-3)}{(n+3)^2(n-3)} - \frac{n+3}{(n-3)(n+3)^2}\]
05
Combine the Fractions
Since the denominators are now the same, subtract the numerators: \[\frac{4(n-3) - (n+3)}{(n+3)^2(n-3)}\]
06
Simplify the Numerator
Distribute and combine like terms in the numerator: \[4(n-3) - (n+3) = 4n - 12 - n - 3 = 3n - 15\].So the expression becomes: \[\frac{3n - 15}{(n+3)^2(n-3)}\]
07
Factor the Numerator
Factor out the common factor in the numerator (3): \[\frac{3(n-5)}{(n+3)^2(n-3)}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
factoring polynomials
When we simplify fractions with polynomials, we often need to factor the polynomials. Factoring means breaking down a complex polynomial into simpler parts or products of other polynomials.
For example, the polynomial \( n^2 + 6n + 9 \) can be factored into \( (n + 3)^2 \). This is because \( (n + 3) \cdot (n + 3) = n^2 + 6n + 9 \).
Similarly, the polynomial \( n^2 - 9 \) can be factored using the difference of squares method, becoming \( (n - 3)(n + 3) \), since \( (a + b)(a - b) = a^2 - b^2 \).
Factoring polynomials simplifies the expression and makes further steps easier. It's like finding the building blocks of the polynomial.
For example, the polynomial \( n^2 + 6n + 9 \) can be factored into \( (n + 3)^2 \). This is because \( (n + 3) \cdot (n + 3) = n^2 + 6n + 9 \).
Similarly, the polynomial \( n^2 - 9 \) can be factored using the difference of squares method, becoming \( (n - 3)(n + 3) \), since \( (a + b)(a - b) = a^2 - b^2 \).
Factoring polynomials simplifies the expression and makes further steps easier. It's like finding the building blocks of the polynomial.
common denominators
When dealing with fractions, we need a common denominator to combine them. The common denominator is the least common multiple (LCM) of the denominators.
For the fractions \( \frac{4}{(n+3)^2} \) and \( \frac{1}{(n-3)(n+3)} \), the common denominator combines both denominators: \( (n+3)^2(n-3) \).
By rewriting each fraction with this common denominator, we facilitate the subtraction of the fractions. It's important to ensure both fractions are expressed in terms of this LCM to combine their numerators effectively.
For the fractions \( \frac{4}{(n+3)^2} \) and \( \frac{1}{(n-3)(n+3)} \), the common denominator combines both denominators: \( (n+3)^2(n-3) \).
By rewriting each fraction with this common denominator, we facilitate the subtraction of the fractions. It's important to ensure both fractions are expressed in terms of this LCM to combine their numerators effectively.
difference of squares
The difference of squares is a special algebraic expression of the form \( a^2 - b^2 \). It can always be factored into \( (a - b)(a + b) \).
In our example, \( n^2 - 9 \) is a difference of squares. Here, \( a = n \) and \( b = 3 \), so this becomes \( (n - 3)(n + 3) \).
Recognizing and applying the difference of squares makes factoring quick and straightforward. This method simplifies seemingly complicated polynomials by breaking them into simpler binomials.
In our example, \( n^2 - 9 \) is a difference of squares. Here, \( a = n \) and \( b = 3 \), so this becomes \( (n - 3)(n + 3) \).
Recognizing and applying the difference of squares makes factoring quick and straightforward. This method simplifies seemingly complicated polynomials by breaking them into simpler binomials.
numerator and denominator
In any fraction, the top part is the numerator and the bottom part is the denominator. Simplifying the fraction involves both parts.
For our example, the numerator goes through subtraction and simplification once a common denominator is found. For instance, the numerator \( 4(n - 3) - (n + 3) \) simplifies to \( 4n - 12 - n - 3 = 3n - 15 \).
Then, we factor the numerator further, getting \( 3(n - 5) \). The final step combines this numerator with the simplified common denominator for our final simplified fraction. It's crucial to simplify both parts to their simplest forms for less complex expressions.
For our example, the numerator goes through subtraction and simplification once a common denominator is found. For instance, the numerator \( 4(n - 3) - (n + 3) \) simplifies to \( 4n - 12 - n - 3 = 3n - 15 \).
Then, we factor the numerator further, getting \( 3(n - 5) \). The final step combines this numerator with the simplified common denominator for our final simplified fraction. It's crucial to simplify both parts to their simplest forms for less complex expressions.
