Problem 30 - Algebra 2 and Trigonometry

Question

Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. \(\frac{5(1-b)+15}{b^{2}-16}\)

Step-by-Step Solution

Answer

The simplified form is \(-\frac{5}{b+4}\), undefined for \(b=4\) and \(b=-4\).

Step 1: Simplify the Numerator

The numerator is given as \(5(1-b) + 15\). First, distribute the 5: \(5 \cdot 1 - 5 \cdot b = 5 - 5b\). This simplifies to \(5 - 5b + 15\), and then combine like terms: \(20 - 5b\).

Step 2: Simplify the Denominator

The denominator is \(b^2 - 16\). This is a difference of squares, which can be factored as \((b-4)(b+4)\).

Step 3: Combine and Simplify the Fraction

The expression so far is \(\frac{20 - 5b}{(b-4)(b+4)}\). Notice that the numerator \(20 - 5b\) can be factored by taking out a common factor of 5: \(5(4-b)\), which rearranges to \(-5(b-4)\). Thus, the fraction becomes \(-\frac{5(b-4)}{(b-4)(b+4)}\).

Step 4: Cancel Common Factors

Since \(b-4\) is a common factor in the numerator and denominator, they cancel out with each other, leaving \(-\frac{5}{b+4}\).

Step 5: Identify Undefined Values

The original fraction is undefined when the denominator is zero. Set \((b-4)(b+4) = 0\); thus, \(b=4\) and \(b=-4\) make the denominator zero, so the expression is undefined for these values.

Key Concepts

Factoring Polynomials

Polynomials are mathematical expressions consisting of variables and coefficients. Factoring polynomials is a process of breaking them down into simpler components, called factors, which are multiplied together to form the original polynomial.
In our exercise, the first step is to simplify the numerator, which starts as \(5(1-b) + 15\). We distribute the 5 to both terms inside the parenthesis, resulting in \(5 - 5b\). Combining this with the constant 15 gives us \(20 - 5b\).

Factoring can also involve pulling out common factors. Here, the numerator \(20 - 5b\) can be factored by taking out the greatest common factor of 5, yielding \(5(4-b)\). This process makes it easier to identify common terms with the denominator, which helps in simplifying the whole expression.

Undefined Values

Rational expressions become undefined when their denominators equal zero.
To find these undefined values, you need to determine when this problematic condition occurs.

• Begin by setting the denominator equal to zero and solving for the variable.
• In our case, the denominator is \(b^2 - 16\).

The polynomial \(b^2 - 16\) factors to \((b-4)(b+4)\). Setting these equal to zero gives:

• \(b - 4 = 0 \) or \(b = 4\)
• \(b + 4 = 0 \) or \(b = -4\)

These calculated values \(b = 4\) and \(b = -4\) are special because if substituted back into the original rational expression, they would make the denominator zero, rendering the expression undefined. Make sure to note these undefined values when simplifying rational expressions.

Difference of Squares

The difference of squares is a specific type of polynomial that can be factored into two conjugate binomials. It takes the form \(a^2 - b^2\), which can be broken down as \((a-b)(a+b)\). This pattern is crucial for factoring purposes.
In the exercise, the denominator \(b^2 - 16\) is a difference of squares because it can be expressed as \((b)^2 - (4)^2\). Following the difference of squares formula, we factor it into \((b-4)(b+4)\).

Recognizing and applying the difference of squares allows us to simplify complex expressions efficiently. This powerful factoring technique not only aids in simplification but also is instrumental in identifying undefined values that occur when these factors equate to zero.