Problem 78
Question
For exercises \(75-78\), one part of simplifying a rational expression is completed. Problem: To simplify \(\frac{\frac{x^{2}-2 x-15}{x^{2}-7 x+10}}{\frac{x^{2}+2 x-8}{x^{2}+2 x-3}}\), the first step is to factor the polynomials completely. Incorrect Answer: $$ \begin{array}{r} \frac{\frac{x^{2}-2 x-15}{x^{2}-7 x+10}}{\frac{x^{2}+2 x-8}{x^{2}+2 x-3}} \\ =\frac{\frac{(x-5)(x+3)}{(x-2)(x-5)}}{\frac{(x+2)(x-4)}{(x+3)(x-1)}} \end{array} $$
Step-by-Step Solution
Verified Answer
Simplify \ \frac{(x + 3)(x - 1)}{(x - 2)(x + 4)} \.
1Step 1 - Factor the numerator and denominator
First, factor all the polynomials in the expression. For the numerator of the first fraction, factor \(x^{2} - 2x - 15\): \ (x - 5)(x + 3)\. For the denominator of the first fraction, factor \(x^{2} - 7x + 10\): \ (x - 5)(x - 2)\. For the numerator of the second fraction, factor \(x^{2} + 2x - 8\): \ (x + 4)(x - 2)\. For the denominator of the second fraction, factor \(x^{2} + 2x - 3\): \ (x + 3)(x - 1)\.
2Step 2 - Rewrite the expression with the factored forms
Replace the factored polynomials in the expression: \[ \frac{\frac{(x - 5)(x + 3)}{(x - 5)(x - 2)}}{\frac{(x + 4)(x - 2)}{(x + 3)(x - 1)}} \]
3Step 3 - Simplify the complex fraction
Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator: \[ \frac{(x - 5)(x + 3)}{(x - 5)(x - 2)} \times \frac{(x + 3)(x - 1)}{(x + 4)(x - 2)} \]
4Step 4 - Cancel common factors and simplify
Cancel any common factors from the numerator and the denominator: \[ \frac{(x + 3)}{(x - 2)} \times \frac{(x - 1)}{(x + 4)} = \frac{(x + 3)(x - 1)}{(x - 2)(x + 4)} \]
Key Concepts
Factoring PolynomialsComplex FractionsCanceling Common FactorsReciprocal of a Fraction
Factoring Polynomials
In mathematics, factoring polynomials means breaking down a polynomial into simpler 'factor' polynomials whose product is the original polynomial. Factoring is a key step in simplifying rational expressions.
To factor a polynomial, look for two binomials that multiply to produce the original polynomial. For example, to factor the quadratic polynomial \(x^2 - 2x - 15\), find two numbers that multiply to -15 and add to -2. These numbers are 3 and -5, so we can write the polynomial as \((x - 5)(x + 3)\).
Factorizing polynomials ensures that we can work with simpler expressions and identify common factors more easily in later steps.
To factor a polynomial, look for two binomials that multiply to produce the original polynomial. For example, to factor the quadratic polynomial \(x^2 - 2x - 15\), find two numbers that multiply to -15 and add to -2. These numbers are 3 and -5, so we can write the polynomial as \((x - 5)(x + 3)\).
Factorizing polynomials ensures that we can work with simpler expressions and identify common factors more easily in later steps.
Complex Fractions
A complex fraction is a fraction where the numerator, or the denominator, or both, contains a fraction themselves. Simplifying complex fractions can look tricky, but the steps are straightforward.
To simplify a complex fraction, you need to compact the fraction into a single simpler fraction. This often involves rewriting the complex fractions as a multiplication problem. In the example problem, we have \frac{\frac{x^2 - 2x - 15}{x^2 - 7x + 10}}{\frac{x^2 + 2x - 8}{x^2 + 2x - 3}}\.
We factor each polynomial first, and then apply the rule \(\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \times \frac{d}{c}\). This helps us transform our complex fraction into a simpler one.
To simplify a complex fraction, you need to compact the fraction into a single simpler fraction. This often involves rewriting the complex fractions as a multiplication problem. In the example problem, we have \frac{\frac{x^2 - 2x - 15}{x^2 - 7x + 10}}{\frac{x^2 + 2x - 8}{x^2 + 2x - 3}}\.
We factor each polynomial first, and then apply the rule \(\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \times \frac{d}{c}\). This helps us transform our complex fraction into a simpler one.
Canceling Common Factors
Canceling common factors is an essential step in simplifying any rational expression. Once you have factored polynomials in both the numerator and the denominator, look for any factors that are the same in both.
For example, in the problem, after factoring the polynomials, we get: \(\frac{(x - 5)(x + 3)}{(x - 5)(x - 2)} \times \frac{(x + 3)(x - 1)}{(x + 4)(x - 2)}\). You can see that both \frac{(x-5)}{(x-5)}\ and \frac{(x-2)}{(x-2)}\ appear in both the numerator and the denominator. These can be canceled out, simplifying your expression to: \(\frac{(x+3)}{(x-2)} \times \frac{(x-1)}{(x+4)}\). Removing common factors gets us closer to the simplest form of the expression.
For example, in the problem, after factoring the polynomials, we get: \(\frac{(x - 5)(x + 3)}{(x - 5)(x - 2)} \times \frac{(x + 3)(x - 1)}{(x + 4)(x - 2)}\). You can see that both \frac{(x-5)}{(x-5)}\ and \frac{(x-2)}{(x-2)}\ appear in both the numerator and the denominator. These can be canceled out, simplifying your expression to: \(\frac{(x+3)}{(x-2)} \times \frac{(x-1)}{(x+4)}\). Removing common factors gets us closer to the simplest form of the expression.
Reciprocal of a Fraction
The reciprocal of a fraction simply means swapping the numerator and the denominator. For example, the reciprocal of \(\frac{a}{b}\) is \(\frac{b}{a}\).
When working with complex fractions, you'll often need to take the reciprocal of the denominator fraction to simplify it. For instance, in the exercise, after factoring, the complex fraction looks like this: \(\frac{(x - 5)(x + 3)}{(x - 2)(x - 5)} \times \frac{(x - 1)(x + 3)}{(x - 2)(x + 4)}\).
Multiplying by the reciprocal of the denominator helps blend two fractions into one simple fraction. It is a key step to eliminate the complex fraction structure.
When working with complex fractions, you'll often need to take the reciprocal of the denominator fraction to simplify it. For instance, in the exercise, after factoring, the complex fraction looks like this: \(\frac{(x - 5)(x + 3)}{(x - 2)(x - 5)} \times \frac{(x - 1)(x + 3)}{(x - 2)(x + 4)}\).
Multiplying by the reciprocal of the denominator helps blend two fractions into one simple fraction. It is a key step to eliminate the complex fraction structure.
Other exercises in this chapter
Problem 77
For exercises 77-86, find any values of the variable for which this expression is undefined. $$ \frac{c-9}{c+3} $$
View solution Problem 78
Find the number of 80,500 children and adolescents who are obese. Round to the nearest hundred. Approximately one in six children and adolescents are obese. (So
View solution Problem 78
For exercises \(55-86\), use prime factorization to find the least common multiple. $$ 168 n^{2} w^{6} ; 252 n^{2} w^{2} $$
View solution Problem 78
For exercises 39-82, simplify. $$ \frac{b^{2}+7 b-18}{b-2} \div \frac{b+9}{b+4} $$
View solution