Problem 100 - College Algebra with Modeling and Visualization

Question

Factor the expression. \(125 r^{6}+64 t^{3}\)

Step-by-Step Solution

Answer

The expression is factored as \((5r^2 + 4t)(25r^4 - 20r^2t + 16t^2)\).

Step 1: Recognize the Sum of Cubes

The expression given, \(125 r^{6} + 64 t^{3}\), can be recognized as a sum of cubes. It is of the form \(a^3 + b^3\), where \(a = (5 r^2)\) and \(b = (4 t)\). This is because \(125 r^6\) can be rewritten as \((5 r^2)^3\), and \(64 t^3\) can be rewritten as \((4 t)^3\).

Step 2: Apply the Sum of Cubes Formula

The formula for factoring the sum of cubes is \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Identifying \(a\) and \(b\) from Step 1, we apply this formula with \(a = 5r^2\) and \(b = 4t\).

Step 3: Plug Values into the Formula

Substitute \(a = 5r^2\) and \(b = 4t\) into the sum of cubes formula:First, find \(a + b\):\[a + b = 5r^2 + 4t\]Then, find \(a^2 - ab + b^2\):\[a^2 = (5r^2)^2 = 25r^4\]\[ab = (5r^2)(4t) = 20r^2t\]\[b^2 = (4t)^2 = 16t^2\]Hence, \[a^2 - ab + b^2 = 25r^4 - 20r^2t + 16t^2\].

Step 4: Final Factored Form

The expression \(125r^6 + 64t^3\) can be fully factored using the sum of cubes formula as:\[(5r^2 + 4t)(25r^4 - 20r^2t + 16t^2)\]

Key Concepts

Sum of Cubes

The sum of cubes is an important concept when dealing with polynomial expressions. It relates to the identity used to factor expressions of the form \(a^3 + b^3\). Understanding this identity is key:

• \(a^3 + b^3\) can always be rewritten as \((a + b)(a^2 - ab + b^2)\)

This formula helps simplify polynomials that appear complex at first. The first step to use this identity is to recognize if a given polynomial actually fits the sum of cubes form. In the exercise above, \(125 r^6 + 64 t^3\) is identified as such:

• \(a = (5r^2)\) because \(125 r^6 = (5r^2)^3\)
• \(b = (4t)\) because \(64 t^3 = (4t)^3\)

Recognizing a sum of cubes allows us to apply the special factoring formula and break the expression down into simpler terms.

Polynomial Expression

A polynomial expression consists of terms that involve variables with non-negative integer exponents. Each term in the expression contributes to the overall complexity and degree of the polynomial. For example, \(125r^6 + 64t^3\) consists of two terms.

• The first term, \(125r^6\), is a single term made of both a constant and a variable raised to a power.
• The second term, \(64t^3\), is also a term with a coefficient and a variable raised to a power.

Knowing how to manipulate polynomial expressions by identifying special types, like the sum of cubes, offers a systematic way to simplify them. By reducing these expressions into simpler forms, students can solve more complex algebraic problems effectively.
Polynomials show up in many math problems, and recognizing their structures and forms is critical for mastering algebra.

Algebraic Manipulation

Algebraic manipulation involves re-writing or transforming mathematical expressions to simplify or solve them. This could mean expanding, factoring, or combining like terms. The process requires attention to formulas and rules that facilitate such transformations.In our step-by-step solution:

• First, we recognize the expression as a sum of cubes form.
• Using the sum of cubes formula \((a^3 + b^3 = (a + b)(a^2 - ab + b^2))\), we substitute known values of \(a\) and \(b\).
• Perform calculations: find \(a + b\), calculate \(a^2\), \(ab\), and \(b^2\), and combine these to get \(a^2 - ab + b^2\)

This breakdown demonstrates how algebraic manipulation translates a perceiveably "difficult" polynomial into simpler, more workable expressions. The manipulations remove complexity by applying the relevant algebraic identities and making calculations manageable.