Problem 82 - Algebra for College Students

Question

Perform the indicated operations. $$\frac{5-10 k}{k^{2}-2 k} \div \frac{2 k^{2}+7 k-4}{k^{2}+2 k-8}$$

Step-by-Step Solution

Answer

\(\frac{-5}{k}\)

Step 1: Simplify the given fractions

Before dividing fractions, simplify the numerators and denominators. Factorize both the numerator and the denominator of each fraction where possible. Let's start by simplifying the first fraction, \( \frac{5-10k}{k^2-2k} \). Factor out the common factor:\[\frac{5(1-2k)}{k(k-2)}\]

Step 2: Simplify the second fraction

Next, simplify the second fraction, \( \frac{2k^2+7k-4}{k^2+2k-8} \). Factorize both numerator and denominator:Numerator: \[2k^2 + 7k - 4 = (2k - 1)(k + 4)\]Denominator: \[k^2 + 2k - 8 = (k + 4)(k - 2)\]Thus, the fraction simplifies to \[\frac{(2k-1)(k+4)}{(k+4)(k-2)}\]

Step 3: Rewrite the division of fractions as multiplication

Rewrite the division of two fractions as multiplication by flipping the second fraction and multiplying:\[\frac{5(1-2k)}{k(k-2)} \times \frac{(k+4)(k-2)}{(2k-1)(k+4)}\]

Step 4: Cancel out the common terms

Cancel any common factors in the numerator and the denominator. We have \((k-2)\) and \((k+4)\) as common factors in the multiplication:\[\frac{5(1-2k)}{k} \times \frac{1}{2k-1}\]

Step 5: Final multiplication

Combine the simplified fractions:\[\frac{5(1-2k)}{k(2k-1)} = \frac{5(2k-1)(-1)}{k(2k-1)} = \frac{-5}{k}\]

Key Concepts

factoring polynomials

Factoring polynomials is a key process in simplifying rational expressions. When we factorize, we break down a complex polynomial into products of simpler polynomials. This is similar to finding the prime factors of a number. Taking an example from the exercise, the denominator in the first fraction is initially presented as \(k^2 - 2k\). To factorize it, look for common factors: \[k^2 - 2k = k(k - 2)\]. Similarly, for the second fraction, the numerator \(2k^2 + 7k - 4\) factors into \((2k - 1)(k + 4)\). Recognizing and performing this step simplifies the computation of the entire expression.

multiplication of fractions

Multiplying fractions involves combining the numerators together and the denominators together. Before doing this, it's important to factorize wherever possible to cancel out common factors later. For example, when multiplying the simplified fractions from our exercise, we have: \(\frac{5(1-2k)}{k(k-2)} \times \frac{(k+4)(k-2)}{(2k-1)(k+4)}\). Multiplying these directly would be tedious, but by canceling out common factors, we reduce our work significantly. This approach not only simplifies our calculations but also reduces the complexity.

division of fractions

Dividing fractions might seem difficult, but it's straightforward if you convert it into multiplication. When dividing by a fraction, you multiply by its reciprocal. The reciprocal of a fraction is achieved by flipping the numerator and denominator. For instance, in our given problem: \(\frac{5-10k}{k^2-2k} \div \frac{2k^2+7k-4}{k^2+2k-8}\), convert it to multiplication: \(\frac{5(1-2k)}{k(k-2)} \times \frac{(k+4)(k-2)}{(2k-1)(k+4)}\). This transforms a division problem into a simpler multiplication problem, making it easier to work through.

simplifying algebraic expressions

Simplifying algebraic expressions is crucial to making computations easier and more understandable. This involves canceling common terms and reducing fractions to their simplest form. Referring back to the exercise, after converting and simplifying the fractions, we had: \(\frac{5(1-2k)}{k(k-2)} \times \frac{(k+4)(k-2)}{(2k-1)(k+4)}\). By canceling the common terms \((k-2)\) and \((k+4)\), we simplified the expression to \(\frac{5(1-2k)}{k} \times \frac{1}{2k-1}\). Further simplification led us to the final reduced form \(\frac{-5}{k}\). Simplifying expressions helps in understanding the essential components of an algebraic problem without unnecessary complexity.