Problem 27 - Elementary and Intermediate Algebra

Question

Factor difference of two squares. \(36 x^{4} y^{2}-49 z^{6}\)

Step-by-Step Solution

Answer

The expression factors to \((6x^2y - 7z^3)(6x^2y + 7z^3)\).

Step 1: Identify the Difference of Squares Format

First, recognize that the expression can be put in the form of \(a^2 - b^2\), which is the format for the difference of two squares. The expression \(36x^4y^2 - 49z^6\) already resembles it, where: \((6x^2y)^2 = 36x^4y^2\) and \((7z^3)^2 = 49z^6\). So, we have \((6x^2y)^2 - (7z^3)^2\).

Step 2: Apply the Difference of Squares Formula

Recall the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = 6x^2y\) and \(b = 7z^3\). Substitute these into the formula to get: \((6x^2y - 7z^3)(6x^2y + 7z^3)\).

Step 3: Verify the Factorization

Expand the factored form to ensure it matches the original expression. Multiply: \((6x^2y - 7z^3)(6x^2y + 7z^3)\). This results in \((6x^2y)^2 - (7z^3)^2 = 36x^4y^2 - 49z^6\), confirming the factorization is correct.

Key Concepts

Difference of Squares

The difference of squares is a specific type of algebraic expression. It involves two perfect squares separated by a subtraction sign. This form can always be rewritten using the identity:

• the expression in the form of \(a^2 - b^2\)
• factored as \((a - b)(a + b)\)

In our example, the expression \(36x^4y^2 - 49z^6\) fits this pattern perfectly. Identifying the parts of this algebraic expression means recognizing that both terms are perfect squares. That's because \((6x^2y)^2 = 36x^4y^2\) and \((7z^3)^2 = 49z^6\).
This understanding allows us to factor the equation quickly and efficiently, reducing complex equations into simpler forms.

Algebraic Expressions

Algebraic expressions consist of numbers, variables, and operations like addition, subtraction, multiplication, and division. Variables represent unknown or changeable values, usually denoted by letters like \(x\), \(y\), and \(z\).
In algebra, understanding how to manipulate these expressions is essential for solving equations and finding answers to complex problems.
In our factorization problem, we work with expression terms such as \(36x^4y^2\) and \(49z^6\). Here:

• Each term is a combination of coefficients (numerical parts like \(36\) and \(49\))
• along with variables raised to a power (like \(x^4\), \(y^2\), and \(z^6\)).

Being comfortable with breaking down and reconstructing these expressions is key for mastering algebra.

Factorization Techniques

Factorization is the process of breaking down an algebraic expression into a product of simpler expressions.
It's like finding components that multiply to give you the original expression. There are various techniques, but some key factorization methods include:

• Finding common factors
• Using the difference of squares formula
• Applying special formulas like perfect square trinomials or sum/difference of cubes

Our problem used the difference of squares method. This special technique is particularly useful and straightforward because it helps to split an expression into a product of two binomials: \((a - b)(a + b)\).
This not only simplifies complex polynomial expressions but also helps in solving equations where the expression equals zero, setting each factor to zero separately to find potential solutions.
Practicing these techniques enhances your ability to solve various algebraic problems efficiently.