Problem 67 - Algebra A Combined Function

Question

Factor. $$ 125 y^{3}-8 x^{3} $$

Step-by-Step Solution

Answer

\((5y - 2x)(25y^2 + 10xy + 4x^2)\)

Step 1: Recognize the Expression Type

The given expression is a difference of cubes, as each term is a perfect cube: \(125y^3 = (5y)^3\) and \(8x^3 = (2x)^3\). The expression can be rewritten as \((5y)^3 - (2x)^3\).

Step 2: Apply the Difference of Cubes Formula

Recall the formula for the difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Here, \(a = 5y\) and \(b = 2x\). Plug these values into the formula.

Step 3: Substitute and Simplify

Substitute \(a = 5y\) and \(b = 2x\) into the formula: \((5y)^3 - (2x)^3 = (5y - 2x)((5y)^2 + (5y)(2x) + (2x)^2)\). Simplify each term: - \((5y)^2 = 25y^2\)- \((5y)(2x) = 10xy\) - \((2x)^2 = 4x^2\)

Step 4: Assemble the Factored Form

Combine the simplified terms into the expression inside the second set of parentheses: \((5y - 2x)(25y^2 + 10xy + 4x^2)\). This is the completely factored form of the original expression.

Key Concepts

Difference of Cubes

In algebra, expressions like the one you've just seen—where each term is a perfect cube—are frequently encountered. These are known as a "difference of cubes". Recognizing them makes factoring much easier.

The difference of cubes formula provides a straightforward path to factor such expressions. The formula is:

• If you have an expression of the form \( a^3 - b^3 \), it can be factored as \( (a - b)(a^2 + ab + b^2) \).

This is powerful because it simplifies expressions that look complex. For the given problem, it helps us work with large numbers by breaking them into manageable parts: \( 125y^3 \) becomes \( (5y)^3 \) and \( 8x^3 \) becomes \( (2x)^3 \). Recognizing this pattern will improve your factoring skills dramatically.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operators (such as addition or multiplication). The expression in this exercise is an example of an algebraic expression.

• Algebraic expressions can often be simplified or factored to make them easier to work with.

Understanding the parts of an algebraic expression and practicing how to manipulate them are essential skills in algebra. In our specific case, knowing how to express terms like \( 125y^3 \) and \( 8x^3 \) as cubes allows us to simplify the expression further.

Moreover, by comprehending the operations involved, such as using the formula for the difference of cubes, you can solve problems more efficiently and gain deeper insights into algebraic manipulation.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into its simplest parts, or "factors". These factors are simpler polynomials, and when multiplied together, they return the original polynomial.

For expressions that are a difference of cubes, such as \( 125y^3 - 8x^3 \), using the difference of cubes formula is a reliable method to achieve this break-down. Once we determine that \( a = 5y \) and \( b = 2x \), we substitute into the formula:

• \( (5y - 2x) \), capturing the linear difference between the two bases.
• \( (25y^2 + 10xy + 4x^2) \), constructing the "quadratic" expression involving both terms.

This form of factorization uncovers the underlying structure of the polynomial, making it more manageable and easier to understand. By regularly practicing this method, you'll become adept at transforming complex expressions into easily solvable problems.