Problem 13 - Intermediate Algebra

Question

Factor each denominator completely. a. \(\frac{17}{40 x^{2}}\) b. \(\frac{x+25}{2 x^{2}-6 x}\) c. \(\frac{n^{2}+3 n-4}{n^{2}-64}\)

Step-by-Step Solution

Answer

a. \(2^3 \cdot 5 \cdot x^2\), b. \(2x(x - 3)\), c. \((n - 8)(n + 8)\).

Step 1: Factor the Denominator for Part (a)

The denominator of the first expression is \(40x^2\). We can factor this as follows:1. Recognize that the number 40 can be factored into prime factors: \(40 = 2^3 \times 5\).2. Combine this with \(x^2\) to get the complete factorization: \(40x^2 = 2^3 \times 5 \times x^2\).Thus, the complete factorization is \(2^3 \cdot 5 \cdot x^2\).

Step 2: Factor the Denominator for Part (b)

The denominator is \(2x^2 - 6x\). To factor this expression:1. Identify the greatest common factor (GCF) of the terms, which is \(2x\).2. Factor out \(2x\) from each term: \(2x^2 - 6x = 2x(x - 3)\).Thus, the complete factorization is \(2x(x - 3)\).

Step 3: Factor the Denominator for Part (c)

The denominator is \(n^2 - 64\) which is a difference of squares. Recognize that:1. \(n^2\) is a perfect square: \((n)^2\).2. 64 is also a perfect square: \((8)^2\).3. Apply the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\).Thus, \(n^2 - 64 = (n - 8)(n + 8)\).

Key Concepts

Greatest Common Factor

When dealing with polynomials, one of the first steps in factoring is to identify the greatest common factor (GCF). The GCF is the largest factor shared by all terms in a polynomial. By identifying and factoring out the GCF, we simplify the polynomial and make further factoring easier.

Here's how to find the GCF:

• List out the factors of each term in the polynomial. For example, in the polynomial \(2x^2 - 6x\), the terms are \(2x^2\) and \(-6x\).
• The factors of \(2x^2\) are \(2\) and \(x^2\), while the factors of \(-6x\) are \(-1\), \(2\), \(3\), and \(x\).
• The largest factor common to both terms is \(2x\).
• Pull out this GCF: \(2x(x - 3)\).

By factoring out the GCF, you are simplifying the polynomial and preparing it for further factoring if needed. This technique makes solving equations quicker and reduces the complexity of the polynomials you are working to simplify or solve.

Difference of Squares

The difference of squares is a specific type of polynomial where two perfect squares are subtracted from each other. The formula to remember is: \[a^2 - b^2 = (a - b)(a + b)\]This formula is incredibly useful for factoring polynomials like \(n^2 - 64\).

Here's the process:

• Identify the perfect squares. For \(n^2 - 64\), notice that \(n^2\) is the square of \(n\), and \(64\) is the square of \(8\).
• Using the difference of squares formula, rewrite the expression as the product of two binomials: \((n - 8)(n + 8)\).

This method quickly breaks down a polynomial into its simpler components, making it easier to factor or even to solve. Recognizing a difference of squares is a key skill in algebra that leads to efficient problem-solving.

Prime Factorization

Prime factorization involves breaking down numbers into a product of their prime numbers, which are numbers greater than 1 that have no divisors other than 1 and themselves. This technique is a cornerstone in simplifying numbers and expressions, particularly when factoring polynomials.

To perform prime factorization, follow these steps:

• For a number like 40, begin by dividing by the smallest prime, 2: \(40 \div 2 = 20\), then \(20 \div 2 = 10\), and \(10 \div 2 = 5\), where 5 is a prime number.
• This division gives the prime factorization: \(40 = 2^3 \times 5\).
• For polynomial terms, incorporate the variable factor with primes, like \(x^2\) in \(40x^2\).
• The full factorization of \(40x^2\) becomes \(2^3 \times 5 \times x^2\).

Using prime factorization allows you to present complex numeric expressions in a simpler, unified form, which aids in further algebraic manipulation.