Problem 36 - A Graphical Approach to College Algebra

Question

Find each product or quotient. $$\frac{8 y^{3}-125}{4 y^{2}-20 y+25} \cdot \frac{2 y-5}{y}$$

Step-by-Step Solution

Answer

The expression simplifies to \(\frac{4y^2 + 10y + 25}{y}\).

Step 1: Factor the Numerator and Denominator

To simplify the expression, first factor the numerator and denominator of the rational expressions. The term \(8y^3 - 125\) is the difference of cubes, which factors into \((2y - 5)(4y^2 + 10y + 25)\). The term \(4y^2 - 20y + 25\) is a perfect square that can be rewritten as \((2y - 5)^2\).

Step 2: Simplify the First Fraction

The first fraction \(\frac{8y^3 - 125}{4y^2 - 20y + 25}\) simplifies to \(\frac{(2y - 5)(4y^2 + 10y + 25)}{(2y - 5)(2y - 5)}\). Cancel out the common factor \(2y - 5\) in the numerator and the denominator to obtain \(\frac{4y^2 + 10y + 25}{2y - 5}\).

Step 3: Simplify the Whole Expression

Now simplify the whole expression by multiplying the simplified first fraction by the second fraction, \(\frac{2y - 5}{y}\). The expression becomes \(\frac{4y^2 + 10y + 25}{2y - 5} \times \frac{2y - 5}{y}\).

Step 4: Multiply and Cancel Common Factors

Since the fractions are multiplied, cancel the common factor \(2y - 5\) in both the numerator of the second fraction and the denominator of the first fraction. This leaves us with \(\frac{4y^2 + 10y + 25}{y}\).

Step 5: Final Product or Quotient

The expression now simplifies fully to \(\frac{4y^2 + 10y + 25}{y}\).

Key Concepts

Difference of Cubes

When you encounter the term "difference of cubes," it refers to expressions like \(a^3 - b^3\). This specific format can be factored using the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). In our exercise, the expression \(8y^3 - 125\) is an example of a difference of cubes. Here, \(8y^3\) is written as \((2y)^3\) and \(125\) as \(5^3\). Applying the formula, \(8y^3 - 125\) can be factored into \((2y - 5)(4y^2 + 10y + 25)\). This approach simplifies complex polynomials by breaking them down into manageable factors.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of its factors. This is a crucial step in simplifying expressions like the one in our exercise. Factoring makes expressions easier to manipulate and solve. For instance, in our problem, the numerator \(8y^3 - 125\) was factored as a difference of cubes. Similarly, the denominator \(4y^2 - 20y + 25\) was identified as a perfect square trinomial. By recognizing structured forms, like perfect squares or differences of cubes, you can quickly factor polynomials.

• This requires identifying patterns and using specific formulas or techniques for different forms.
• Factoring simplifies expressions and makes it easier to identify common factors that can be cancelled out.

This process is fundamental whenever simplifying rational expressions.

Simplification

Simplification of rational expressions involves reducing the expression to its simplest form. After factoring both the numerator and the denominator in a rational expression, simplification becomes possible. In our exercise, after factoring, we noticed common terms like \((2y - 5)\) in both the numerator and the denominator. We can "cancel" these terms—meaning we remove identical terms from both the numerator and the denominator, thus simplifying the fraction. This is a critical step as it reduces the complexity of the expression, making further operations easier to perform.

• It often involves canceling common factors.
• The goal is to make the expression as simple as possible for easier computation or evaluation.

Simplification is an essential skill in managing rational expressions efficiently.

Perfect Squares

Perfect squares are quadratic expressions of the form \(a^2 - 2ab + b^2 = (a - b)^2\) or \(a^2 + 2ab + b^2 = (a + b)^2\). In our exercise, the denominator \(4y^2 - 20y + 25\) is a perfect square. It can be rewritten as \((2y - 5)^2\). Recognizing perfect square trinomials is helpful because it allows expressions to be factored quickly using predictable patterns.

• This recognition helps to simplify expressions by revealing opportunities to cancel or replace terms.
• It is a particularly useful technique when dealing with quadratic equations and rational expressions.

Understanding perfect squares improves your ability to manipulate algebraic expressions and solve equations efficiently.