Problem 56 - College Algebra

Question

For the following exercises, factor the polynomials completely. $$ 81 y^{4}-256 $$

Step-by-Step Solution

Answer

\((3y - 4)(3y + 4)(9y^2 + 16)\)

Step 1: Identify the Structure

The given polynomial is \(81y^4 - 256\). This expression is a difference of squares. Recall that a difference of squares is in the form \(a^2 - b^2\) and can be factored into \((a - b)(a + b)\). In this case, recognize that \(81y^4\) is \((9y^2)^2\) and \(256\) is \(16^2\).

Step 2: Apply Difference of Squares Formula

Use the difference of squares formula: for \(a^2 - b^2\), it factors into \((a - b)(a + b)\). Here, \(a = 9y^2\) and \(b = 16\), so it factors to \((9y^2 - 16)(9y^2 + 16)\).

Step 3: Further Factor the Difference of Squares

The term \(9y^2 - 16\) is again a difference of squares because \(9y^2 = (3y)^2\) and \(16 = 4^2\). So, apply the formula: \(9y^2 - 16 = (3y - 4)(3y + 4)\).

Step 4: Check for Additional Factoring

The term \(9y^2 + 16\) is not a difference of squares and cannot be factored further using real numbers. It remains as is.

Key Concepts

Difference of Squares

In mathematics, a difference of squares is a specific type of polynomial that is notably easy to factor. It follows a strict form: \( a^2 - b^2 \). You may recognize this as something you can break down into simpler components using the formula: \( (a-b)(a+b) \). This is incredibly efficient in simplifying expressions, allowing you to transform them into more understandable pieces.

In the context of the problem given, we identify \( 81y^4 - 256 \) as a difference of squares because:

• \( 81y^4 = (9y^2)^2 \)
• \( 256 = 16^2 \)

This gives us two perfect squares, allowing direct application of the difference of squares formula.Ultimately, this factorization simplifies to \( (9y^2 - 16)(9y^2 + 16) \), breaking the original expression into smaller segments, easing further examination or operations.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into a product of its factors. This involves expressing the polynomial in a different form, which can simplify solving equations or further manipulation. Recognizing patterns, such as the difference of squares, assists greatly in this task.

Let's break it down for the given polynomial, \( 81y^4 - 256 \):

• Identify it as a difference of squares: \( (9y^2)^2 - 16^2 \).
• Apply the difference of squares formula to obtain \((9y^2 - 16)(9y^2 + 16)\).
• Notice that \( 9y^2 - 16 \) is also a difference of squares: \( (3y)^2 - 4^2 \).

So, you can further factor it into \( (3y-4)(3y+4) \). The factor \( 9y^2 + 16 \) does not simplify further with real numbers. This multi-step approach is typical in polynomial factorization, revealing a simpler perspective of possibly complex expressions.

Algebraic Expressions

An algebraic expression is a mathematical phrase that includes numbers, variables, and operation symbols. Understanding how to manipulate these expressions is crucial for tackling higher-level math problems.

For our exercise, we are given the algebraic expression \( 81y^4 - 256 \). This task required us to:

• Identify the form of the expression, specifically its difference of squares characteristic.
• Apply algebraic rules (difference of squares formula) to break it down into easier components: \( (9y^2 - 16)(9y^2 + 16) \).
• Further break down components like \( 9y^2 - 16 \) using the same principle.

Remember, identifying forms like a difference of squares in algebraic expressions provides a pathway to simpler solutions, illustrating the power of algebra in problem-solving.