Problem 46
Question
When you add rational expressions, you may need to factor a trinomial to find the LCD. Study the sample below. Then simplify the expressions in Exercises 46–49. $$\text { Sample: } \frac{2 x}{x^{2}-1}+\frac{3}{x^{2}+x-2}=\frac{2 x}{(x+1)(x-1)}+\frac{3}{(x-1)(x+2)}$$ The LCD is \((x+1)(x-1)(x+2)\) Note: If you just used \(\left(x^{2}-1\right)\left(x^{2}+x-2\right)\) as the common denominator, the factor \((x-1)\) would be included twice. $$\frac{2}{x-3}+\frac{x}{x^{2}+3 x-18}$$
Step-by-Step Solution
Verified Answer
The simplified expression is \(\frac{3(x+4)}{(x-3)(x+6)}\)
1Step 1: Factor the Denominators
Start by factoring the denominators in the given expression. In this scenario, \(x^{2}+3x-18\) can be factored into \((x-3)(x+6)\). So the expression becomes: \(\frac{2}{x-3}+\frac{x}{(x-3)(x+6)}\)
2Step 2: Build the LCD
Next, identify the LCD of the denominators. As both fractions have \(x-3\) as a common factor in the denominator, the LCD is \((x-3)(x+6)\)
3Step 3: Rewrite the Fraction
You can then rewrite the first fraction as \(\frac{2(x+6)}{(x-3)(x+6)}\); this is done so that both fractions have the same denominator (the LCD). The expression becomes: \(\frac{2(x+6)}{(x-3)(x+6)}+\frac{x}{(x-3)(x+6)}\)
4Step 4: Simplify Expression
Combine the numerators on top of the common denominator. The expression simplifies to \(\frac{2x+12+x}{(x-3)(x+6)}\) which simplifies further to \(\frac{3x+12}{(x-3)(x+6)}\) or \(\frac{3(x+4)}{(x-3)(x+6)}\) when factoring 3 out of the numerator
Key Concepts
Factoring TrinomialsLeast Common Denominator (LCD)Simplifying Expressions
Factoring Trinomials
When working with rational expressions, it often involves factoring trinomials. Trinomials are algebraic expressions that consist of three terms. A common form of a trinomial is:
- A quadratic trinomial: ax² + bx + c, where a, b, and c are constants.
Least Common Denominator (LCD)
The least common denominator (LCD) is essential when adding or subtracting rational expressions, similar to finding a common denominator in fractions. It is the smallest expression that each denominator can divide into without leaving a remainder.To find the LCD, factor each denominator fully and take the highest power of each unique factor present. This ensures all original denominators can divide into the LCD.For the expression:\[\frac{2}{x-3}+\frac{x}{(x-3)(x+6)}\]The LCD must include all the factors present in the denominators. The denominators are already factored as \((x-3)\) and \((x-3)(x+6)\), so the LCD is:\[(x-3)(x+6)\]Checking each term:
- The first term, \(\frac{2}{x-3}\), needs \((x+6)\) in its denominator to match the LCD.
- The second term already has the LCD.
Simplifying Expressions
Simplifying expressions involves rewriting them in their most concise form. It ensures algebraic expressions are easy to interpret, which is why it's a crucial step in solving rational expressions.After identifying the LCD and modifying each fraction to have a common denominator, the next task is to combine the numerators. For example, consider the rational expression:\[\frac{2(x+6)}{(x-3)(x+6)} + \frac{x}{(x-3)(x+6)}\]Due to the common denominator \((x-3)(x+6)\), we can combine the numerators:\[2(x+6)+x = 2x+12+x\]This simplifies further to:\[3x+12\]To make the expression simpler, check if there are common factors in the numerator and the denominator that can be canceled out.\[\frac{3(x+4)}{(x-3)(x+6)}\]The expression is streamlined into its simplest version. Always look for opportunities to factor and cancel out similar terms. This practice makes calculations easier and solutions more manageable.
Other exercises in this chapter
Problem 45
Which product equals the quotient \((2 x+2) \div \frac{x^{2}+x}{4} ?\) (A) \(\frac{1}{2 x+2} \cdot \frac{x^{2}+x}{4}\) (B) \(\frac{2 x+2}{1} \cdot \frac{x^{2}+x
View solution Problem 45
Divide. Divide \(c^{2}-7 c+21\) by \(2 c-6\)
View solution Problem 46
Simplify. $$\left(-\frac{3}{4}\right)\left(\frac{3 y}{-5}\right)$$
View solution Problem 46
Graph the function. Describe the domain. $$y=\frac{6}{x+9}-7$$
View solution