Problem 21 - Precalculus: Functions and Graphs

Question

Factor the polynomial. $$64 x^{3}-y^{6}$$

Step-by-Step Solution

Answer

The polynomial \(64x^3 - y^6\) factors to \((4x - y^2)(16x^2 + 4xy^2 + y^4)\).

Step 1: Identify the Form of the Polynomial

Recognize that the given polynomial is a difference of two terms: \(64x^3 - y^6\). Notice that \(64x^3\) is a perfect cube \((4x)^3\) and \(y^6\) is also a perfect square as well as a perfect cube \((y^2)^3\).

Step 2: Apply the Difference of Cubes Formula

Recall the difference of cubes formula: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). In this case, let \(a = (4x)\) and \(b = (y^2)\).

Step 3: Substitute and Expand

Substitute the values of \(a\) and \(b\) into the formula. The expression becomes \((4x - y^2)((4x)^2 + (4x)(y^2) + (y^2)^2)\).

Step 4: Simplify Each Term

Simplify each term in the expanded expression: - The first term is \((4x)^2 = 16x^2\).- The second term is \((4x)(y^2) = 4xy^2\).- The third term is \((y^2)^2 = y^4\).

Step 5: Write the Final Factorization

Combine these simplified terms to get the final factorized form: \((4x - y^2)(16x^2 + 4xy^2 + y^4)\).

Key Concepts

Difference of Cubes

When approaching algebraic expressions involving subtraction, spotting the "difference of cubes" can simplify your problem-solving process. The idea centers around two cubes being subtracted in a polynomial.

• Recognize it as in the form of \(a^3 - b^3\).
• This form can be factored using the formula: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\).
• This method reduces complex expressions into simpler factors, making solving easier.

By identifying parts of the polynomial as cubes, like \(64x^3\) as \((4x)^3\) and \(y^6\) as \((y^2)^3\), you can apply the formula directly and efficiently tackle what seems like a daunting algebraic task.

Perfect Cube

A crucial step in recognizing a "difference of cubes" is identifying what constitutes a perfect cube.

• A perfect cube is a number or expression that can be written as another number raised to the power of three.
• Examples include \(8 = 2^3\) and \(27 = 3^3\).
• In algebraic form, \(a^3\) or \((variable)^3\) implies a perfect cube.

In the exercise, both terms \(64x^3\) and \(y^6\) are converted into cubes. Here, \((4x)^3\) perfectly matches the format, allowing straightforward application of the difference of cubes formula. Recognizing and rewriting terms as cubes is a critical skill in algebra simplification.

Algebraic Expression

An algebraic expression like \(64x^3 - y^6\) consists of variables and constants combined using arithmetic operations. Understanding how to manipulate such expressions is foundational in algebra.

• Components include terms, coefficients, and operations such as addition and subtraction.
• Simplifying these expressions involves factoring them into their basic building blocks.
• This can make solving equations more manageable and unveiling underlying patterns or behaviors.

Rewriting complex expressions through factorization not only offers a simplified perspective but aides in deeper conceptual understanding, particularly in polynomials. The exercise above showcases these strategies in action by breaking down the expression into more digestible parts.

Factoring Techniques

Mastering different factoring techniques makes handling complex algebraic polynomials much easier. Common methods include recognizing special products like the difference of cubes.

• Factoring turns a polynomial into a product of simpler expressions.
• For \(64x^3 - y^6\), know the shape: it can be split into \((4x - y^2)(16x^2 + 4xy^2 + y^4)\).
• Such techniques transform intractable problems into solvable ones, offering a clearer path to the solution.

Learning these skills creates a robust mathematical toolkit. By applying form-specific strategies, students can decode seemingly complex equations and gain confidence in manipulating algebraic expressions.