Problem 36 - Elementary and Intermediate Algebra

Question

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 125 p^{3}-64 y^{3} $$

Step-by-Step Solution

Answer

The expression is factored as \((5p - 4y)(25p^2 + 20py + 16y^2)\).

Step 1: Identify the expression as a sum or difference of cubes

The given expression is \(125p^3 - 64y^3\). This appears to be a difference of cubes, since both 125 and 64 are perfect cubes. \(125p^3\) is \((5p)^3\) and \(64y^3\) is \((4y)^3\).

Step 2: Apply the Difference of Cubes Formula

The difference of cubes formula is \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Using \((5p)^3 - (4y)^3\), we can identify \(a = 5p\) and \(b = 4y\).

Step 3: Substitute the values into the formula

Substitute \(a = 5p\) and \(b = 4y\) into the formula: \((5p - 4y)((5p)^2 + (5p)(4y) + (4y)^2)\).

Step 4: Simplify the expression

Calculate each term inside the parentheses: - \((5p)^2 = 25p^2\)- \((5p)(4y) = 20py\)- \((4y)^2 = 16y^2\)So, the factored expression becomes \((5p - 4y)(25p^2 + 20py + 16y^2)\).

Key Concepts

Difference of Cubes

When you come across an expression like the one in the exercise, where subtraction is involved between two cubic terms, you might be dealing with the "Difference of Cubes." This is a specific type of algebraic identity that lets us factor certain expressions neatly.
So, what qualifies an expression as a difference of cubes? Simply, it should take the form of:

• \(a^3 - b^3\)

Using this pattern, you can transform complex cubic polynomials into more manageable products of simpler polynomials.
The magical formula for factoring the difference of cubes is:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

This formula helps break down the expression into a difference and a trinomial. It's a handy tool for simplifying your work with cubic expressions. In our example, it was used effectively to factor \((5p)^3 - (4y)^3\). Just identify your \(a\) and \(b\), plug them into the formula, and voilà! You've got your expression in a factorized form.

Perfect Cubes

In algebra, recognizing perfect cubes is an important skill. A perfect cube is simply a number that can be expressed as something raised to the third power. In simpler terms, it looks like \(n^3\), where \(n\) is a whole number or a more complex term like \(5p\).
To figure out if a number is a perfect cube, ask yourself if there exists a whole number that when cubed, results in that original number.
For example:

• 125 is a perfect cube because it equals \(5^3\).
• Similarly, 64 is a perfect cube since it equals \(4^3\).

Recognizing these can be very helpful, especially when dealing with factoring problems involving cubes, like the one in the exercise.
Perfect cubes often appear in algebraic expressions and allow us to employ identities like the difference of cubes formula to simplify expressions.

Algebraic Expressions

Algebraic expressions are the building blocks of algebra. They're made up of numbers, variables, and mathematical operations like addition, subtraction, multiplication, and occasionally division.
In the exercise, the algebraic expression \(125p^3 - 64y^3\) contains variables \(p\) and \(y\) and coefficients like 125 and 64.
These expressions come to life when we manipulate them—like when factoring. Factoring is the process of breaking down an expression into a product of simpler components, as shown in the solution steps.
Working with algebraic expressions involves many processes:

• Identifying patterns, such as recognizing cubes or factorable differences.
• Applying identities and formulas, like factoring formulas, to simplify and rearrange expressions.

Understanding these processes makes tackling more complex algebraic problems much simpler. Start by getting comfortable with basic expressions, and as you gain more experience, moving to more complex ones will become a natural progression!