Problem 53 - Elementary Algebra

Question

Factor completely. $$ 8 x 3-125 $$

Step-by-Step Solution

Answer

Factor completely: \((2x - 5)(4x^2 + 10x + 25)\).

Step 1: Identify the Type of Expression

The expression given is a polynomial of the form \(8x^3 - 125\). We notice that it fits the structure of a difference of cubes, which is a^3 - b^3.

Step 2: Express as Difference of Cubes

Recognize \(8x^3\) as \((2x)^3\) and \(125\) as \(5^3\). Thus, the expression can be rewritten as a difference of cubes: \((2x)^3 - 5^3\).

Step 3: Apply Difference of Cubes Formula

The difference of cubes formula is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Here, \(a = 2x\) and \(b = 5\). Substitute these into the formula to get: \((2x - 5)((2x)^2 + (2x)(5) + 5^2)\).

Step 4: Simplify the Expression

Simplify the terms inside the second factor: \((2x)^2 = 4x^2\), \((2x)(5) = 10x\), and \(5^2 = 25\). Thus, the expression becomes \((2x - 5)(4x^2 + 10x + 25)\).

Step 5: Verify Factorization

Ensure the factorization is correct by expanding \((2x - 5)(4x^2 + 10x + 25)\) and checking it equals the original expression \(8x^3 - 125\).

Key Concepts

difference of cubes

The concept of the "difference of cubes" applies to polynomials that can be expressed in the form \(a^3 - b^3\). This expression indicates the subtraction ("difference") between two perfect cubes. Recognizing when a polynomial is a difference of cubes is crucial to solving it. In our exercise, \(8x^3 - 125\) is identified as a difference of cubes because both terms are perfect cubes: \(8x^3\) is \((2x)^3\) and 125 is \(5^3\).
Utilizing the difference of cubes formula, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), allows for a seamless approach to factorization. This formula simplifies a potentially complex polynomial into a product of a linear and a quadratic expression.
Hence, it's key to correctly determine \(a\) and \(b\) from the given polynomial. Once identified, plugging them into the formula yields a properly factored expression.

algebraic expressions

Algebraic expressions consist of variables, constants, and operations that all come together to form a meaningful mathematical phrase. In the expression \(8x^3 - 125\), \(8x^3\) involves the variable term \(x\) raised to the power of 3, highlighting its cubic nature. The constant 125 is another crucial element of this polynomial.
Understanding how to manipulate algebraic expressions is essential for tackling problems involving them. Basic operations like addition, subtraction, multiplication, and division within these expressions form the foundation for factoring and simplifying.
With practice, recognizing familiar patterns such as cubes or squares within expressions becomes more intuitive, making factorization and other algebraic operations both easier and faster. The goal is to break down complex expressions into simpler components that are easier to manage.

factoring formulas

Factoring formulas are tools for breaking down expressions into their simplest multiplicative components. For polynomial expressions, factoring simplifies the process of solving equations and finding zeros. The difference of cubes is one significant factoring formula among many others like the sum of cubes, the difference of squares, and the quadratic formula.
The formula for the difference of cubes is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). This formula is versatile, allowing any expression that fits \(a^3 - b^3\) to be factored easily.
Emphasizing the application of the formula: suppose \(a = 2x\) and \(b = 5\), the expression \((2x - 5)((2x)^2 + (2x)(5) + 5^2)\) results after substitution, leading to the correct factorized form. Such formulas are not just methods, but keys to unlocking the simplified forms of polynomials, facilitating further operations like expansion and verification.