Problem 8 - Elementary and Intermediate Algebra

Question

For each of the following polynomials, which factoring method would you use first? $$ y^{3}-64 $$

Step-by-Step Solution

Answer

Use the difference of cubes method to factor \(y^3 - 64\) as \((y - 4)(y^2 + 4y + 16)\).

Step 1: Recognize the Polynomial Type

The given polynomial is \(y^3 - 64\). This expression is a cubic expression of the form \(a^3 - b^3\), where \(a = y\) and \(b^3 = 64\). This suggests it is a difference of cubes.

Step 2: Understand the Difference of Cubes Formula

The difference of cubes can be factored using the formula: \[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\] In this context, \(a = y\) and \(b = 4\) since \(4^3 = 64\).

Step 3: Apply the Difference of Cubes Formula

Substitute \(a = y\) and \(b = 4\) into the formula: \[y^3 - 4^3 = (y - 4)(y^2 + 4y + 16)\]Factor the original polynomial using this method.

Key Concepts

Cubic Polynomial

Polynomials are expressions composed of variables and coefficients, and a cubic polynomial is a special type as it has the highest degree of three. This means the largest exponent of the variable is 3. For example, in the polynomial \( y^3 - 64 \), the term \( y^3 \) indicates it's a cubic polynomial because it contains the cube of the variable \( y \).

Cubic polynomials can appear complex, yet they are often similar to expressions you're already familiar with, especially when dealing with integers. Recognizing them is the first step in learning to factor them.

• The equation \( y^3 - 64 \) clearly represents a cubic polynomial with one single cubic term.
• Understanding cubic polynomials is important as they often showcase symmetry or a pattern which aids in factoring.

Cubic polynomials are often solved by methods like factoring and using formulas specifically designed for handling cubes. It's crucial to identify them correctly to apply the right method.

Difference of Cubes Formula

The difference of cubes is a powerful algebraic tool that helps in factoring expressions of the form \( a^3 - b^3 \). This formula is especially useful in simplifying and solving cubic expressions. For example, the expression \( y^3 - 64 \) can be factored using this method, since it represents the difference between two cubes: \( y^3 \) and \( 4^3 \). Knowing this insight is the key first step in the factorization process.

Here is the formula you will use:

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

Applying this formula helps break down the cubic polynomial into a product of a binomial and a trinomial, which is much simpler to handle. In the current problem:

• Recognize \( a = y \) and \( b = 4 \) since \( 4^3 = 64 \).
• Substituting these values into the formula gives: \((y - 4)(y^2 + 4y + 16)\).

Understanding and using the difference of cubes formula correctly can be very helpful in tackling these algebraic challenges efficiently.

Factoring Techniques

Factoring is the process of breaking down a complex expression into simpler components. For cubic polynomials like \( y^3 - 64 \), the difference of cubes is a prime factoring technique to apply. This involves finding two perfect cubes within a polynomial and simplifying the expression into a product of polynomials.

The application follows three main steps:

• Identify: Recognize the polynomial as a difference of cubes or sum of cubes.
• Formula Application: Apply the respective cubes formula (here, use \( y^3 - 4^3 = (y - 4)(y^2 + 4y + 16) \)).
• Verify: Confirm by expanding or double-checking that the factorization matches the original polynomial.

Factoring reduces expressions into more manageable forms, making them easier to analyze and solve. Intermediate algebra often revisits these techniques because mastery of them opens up further mathematical explorations.