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Chapter 6: Problem 55

Write each rational expression in lowest terms. $$ \frac{2 x^{2}-3 x-5}{2 x^{2}-7 x+5} $$

Short Answer

Expert verified
\(\frac{x + 1}{x - 1}\)

Step by step solution

01

Factor the numerator

Factor the quadratic expression in the numerator. The numerator is \(2x^2 - 3x - 5\). To factor this, look for two numbers that multiply to \(2 \times -5 = -10\) and add up to \(-3\). These numbers are 2 and -5. Rewrite the middle term: \(2x^2 + 2x - 5x - 5\). Factor by grouping: \(2x(x + 1) - 5(x + 1)\). Combine the factored terms: \((2x - 5)(x + 1)\).
02

Factor the denominator

Factor the quadratic expression in the denominator. The denominator is \(2x^2 - 7x + 5\). Look for two numbers that multiply to \(2 \times 5 = 10\) and add up to \(-7\). These numbers are -2 and -5. Rewrite the middle term: \(2x^2 - 2x - 5x + 5\). Factor by grouping: \(2x(x - 1) - 5(x - 1)\). Combine the factored terms: \((2x - 5)(x - 1)\).
03

Simplify the rational expression

Now, place the factored numerator and denominator into the rational expression: \(\frac{(2x - 5)(x + 1)}{(2x - 5)(x - 1)}\). Cancel out the common factor of \(2x - 5\) in the numerator and denominator. The simplified expression is \(\frac{x + 1}{x - 1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
Factoring quadratics is one of the most important steps in simplifying rational expressions. When dealing with quadratic expressions in the form of \(ax^2 + bx + c\), we need to break them down into simpler parts called factors. These factors help us to solve and simplify the expression more easily. Here's a simple way to factor quadratics:

  • Identify two numbers that multiply to \(a \times c\) and add up to \(b\).
  • Rewrite the middle term using these two numbers.
  • Group terms and factor each group separately.
  • Combine the factors to get the fully factored form.

For example, in the numerator \(2x^2 - 3x - 5\), we need two numbers that multiply to \(-10\) and add to \(-3\). These numbers are \(2\) and \(-5\). Rewrite the middle term to get \(2x^2 + 2x - 5x - 5\). Group and factor: \(2x(x + 1) - 5(x + 1)\). Combine to get \((2x - 5)(x + 1)\).
Simplifying Fractions
Simplifying fractions is all about reducing them to their simplest form. When dealing with rational expressions, it's important to factor both the numerator and the denominator first.

After factoring, we can then look for common factors to cancel out. This simplifies the fraction to its lowest terms. In mathematical terms, simplifying fractions involves canceling common factors in both the numerator and the denominator. This makes the expression much simpler and easier to work with.
Canceling Common Factors
Canceling common factors is the final step in simplifying rational expressions. After factoring both the numerator and the denominator, you'll often find common terms in both parts.

To cancel these common factors:
  • Identify the common factors in the numerator and denominator.
  • Cancel these factors from both parts.

In our example, after factoring we got \(\frac{(2x - 5)(x + 1)}{(2x - 5)(x - 1)}\). Notice that \((2x - 5)\) appears in both the numerator and denominator. By canceling out \((2x - 5)\), we're left with \(\frac{x + 1}{x - 1}\). This is the simplified form of the rational expression.

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Most popular questions from this chapter

Write four equivalent forms for each rational expression. $$ -\frac{2 x-9}{7 x-1} $$

Multiply or divide. Write each answer in lowest terms. $$ \frac{6 s^{2}+17 s+10}{s^{2}-4} \cdot \frac{s^{2}-2 s}{6 s^{2}+29 s+20} $$

Multiply or divide. Write each answer in lowest terms. $$ \frac{r^{2}+r-6}{r^{2}+4 r-12} \div \frac{r+3}{r-1} $$

Only one of these choices is equal to \(\frac{\frac{1}{3}+\frac{1}{12}}{\frac{1}{2}+\frac{1}{4}} .\) Which one is it? Answer this question without showing any work, and explain your reasoning. A. \(\frac{5}{9}\) B. \(-\frac{5}{9}\) C. \(-\frac{9}{5}\) D. \(-\frac{1}{12}\)

Multiply or divide. Write each answer in lowest terms. $$ \frac{z^{2}-3 z+2}{z^{2}+4 z+3} \div \frac{z-1}{z+1} $$
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