Problem 89 - Beginning and Intermediate Algebra

Question

Factor completely. $$8 j^{3}+27 k^{3}$$

Step-by-Step Solution

Answer

The completely factored form of the given expression is: \((2j + 3k)(4j^2 - 6jk + 9k^2)\).

Step 1: Identify the sum of cubes

The given expression can be written as a sum of cubes: \(8j^3 + 27k^3 = (2j)^3 + (3k)^3\), where \(a = 2j\) and \(b = 3k\).

Step 2: Apply the sum of cubes formula

Using the formula for factoring a sum of cubes: \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\), we can factor the expression: \((2j)^3 + (3k)^3 = (2j + 3k)((2j)^2 - (2j)(3k) + (3k)^2)\)

Step 3: Simplify the expression

Now, we simplify the expression by expanding the terms inside the parentheses: \((2j + 3k)(4j^2 - 6jk + 9k^2)\) The completely factored form of the given expression is: \((2j + 3k)(4j^2 - 6jk + 9k^2)\).

Key Concepts

Sum of Cubes

In algebra, the **sum of cubes** is a method used to factor certain polynomial expressions. When you have an expression in the form of \( a^3 + b^3 \), it is considered a sum of cubes because both terms are perfect cubes. Recognizing this pattern allows you to apply a specific formula to break down the expression into simpler parts. The formula to factor a sum of cubes is:\[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]This formula tells us how to split the original sum of cubes into two factors:

• The first factor is the sum \( (a + b) \).
• The second factor is a trinomial \( (a^2 - ab + b^2) \), which combines square of \(a\), product of \(a\) and \(b\), and square of \(b\).

For example, given \( 8j^3 + 27k^3 \), we recognize \( (2j)^3 \) and \( (3k)^3 \) as the cubes of \(2j\) and \(3k\), respectively, allowing us to use the sum of cubes formula effectively.

Polynomial Expressions

**Polynomial expressions** are mathematical expressions involving a sum of powers of one or more variables multiplied by coefficients. Each term of a polynomial is made up of a coefficient and a variable raised to a non-negative integer exponent.For example, the expression \(8j^3 + 27k^3\) consists of two terms:

• \(8j^3\), where 8 is the coefficient and \(j^3\) is the term.
• \(27k^3\), where 27 is the coefficient and \(k^3\) is the term.

Polynomials are used extensively in algebra to model real-world phenomena and perform calculations. Each variable in a polynomial can have different powers, and the expression can be manipulated using various algebraic identities and techniques to simplify or solve problems. Recognizing patterns such as sums of cubes within polynomial expressions allows for more straightforward and elegant solutions.

Factoring Techniques

**Factoring techniques** are essential tools in algebra to break down complex expressions into simpler, more manageable parts. The purpose of factoring a polynomial is to express it as a product of its factors, making it easier to work with.A wide range of techniques exist for factoring polynomials, including:

• Greatest common factor (GCF): Identify and factor out the highest common factor shared by the terms in the polynomial.
• Difference of squares: Use the identity \(a^2 - b^2 = (a + b)(a - b)\) to factor expressions with two perfect squares.
• Sum and difference of cubes: Use formulas specifically for expressions like \(a^3 + b^3\) or \(a^3 - b^3\) to factor accordingly.
• Trinomials: Factor expressions of the form \(ax^2 + bx + c\) by finding two numbers that multiply to \(ac\) and add to \(b\).

In the exercise involving \(8j^3 + 27k^3\), identifying it as a sum of cubes allowed us to apply a specialized formula to factor the polynomial expression efficiently. Each technique in factoring aids in simplifying expressions, solving equations, and providing insights into algebraic relationships.