Problem 66 - Elementary and Intermediate Algebra

Question

Factor. $$ 81 y^{4}-100 z^{2} $$

Step-by-Step Solution

Answer

The expression factors to \((9y^2 - 10z)(9y^2 + 10z)\).

Step 1: Recognize the Difference of Squares

The expression \(81y^4 - 100z^2\) is a difference of squares. Remember that a difference of squares is given by the formula \(a^2 - b^2 = (a-b)(a+b)\). Identify \(a\) and \(b\) such that \(a^2 = 81y^4\) and \(b^2 = 100z^2\).

Step 2: Find the Square Roots

Calculate the square roots of \(81y^4\) and \(100z^2\). The square root of \(81y^4\) is \((9y^2)\) since \((9y^2)^2 = 81y^4\). The square root of \(100z^2\) is \((10z)\) since \((10z)^2 = 100z^2\).

Step 3: Apply the Difference of Squares Formula

Using the formula \(a^2 - b^2 = (a-b)(a+b)\), substitute \(a = 9y^2\) and \(b = 10z\). This gives \( (9y^2 - 10z)(9y^2 + 10z) \).

Step 4: Verify the Solution

To verify, expand \((9y^2 - 10z)(9y^2 + 10z)\) using the distributive property: \((9y^2 - 10z)(9y^2 + 10z) = (9y^2)(9y^2) + (9y^2)(10z) - (10z)(9y^2) - (10z)(10z)\). This simplifies to \(81y^4 - 100z^2\), confirming the factorization is correct.

Key Concepts

Difference of Squares

In algebra, the difference of squares is a special case of factoring. It occurs when you have an expression of the form \(a^2 - b^2\). This can be factored into two binomials: \((a - b)(a + b)\). The reason this works is that when you multiply \((a - b)(a + b)\), the middle terms cancel out: \(a^2 + ab - ab - b^2 = a^2 - b^2\).
To identify a difference of squares:

• Ensure the expression has exactly two terms.
• Check if each term is a perfect square.
• Confirm the terms are being subtracted.

In our expression, \(81y^4 - 100z^2\), each term is a perfect square. \(81y^4\) is \((9y^2)^2\) and \(100z^2\) is \((10z)^2\). This means the expression fits the pattern \(a^2 - b^2\), where \(a = 9y^2\) and \(b = 10z\).

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operations. Unlike an equation, it doesn't have an equal sign. The expression \(81y^4 - 100z^2\) is an example because it includes numbers (81 and 100), variables \(y\) and \(z\), and the subtraction operation.
Understanding the parts of an algebraic expression is crucial for manipulation and factorization:

• **Coefficients**: The numbers multiplied by the variables, such as 81 in \(81y^4\).
• **Variables**: Symbols that represent unknown values, \(y\) and \(z\) here.
• **Exponents**: Indicate how many times to multiply the variable by itself, like 4 in \(y^4\).
• **Constants**: Fixed numbers like 81 and 100.

Recognizing these components helps to perform operations and solve for unknowns.

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as a product of its factors. This is an essential skill in algebra as it simplifies complex expressions and makes solving equations easier. The goal is to break down a polynomial into simpler terms that can be easily manipulated or solved.
Steps in factorization include:

• **Identify patterns**: Look for common patterns like the difference of squares, perfect square trinomials, or common factors.
• **Factor terms separately**: Start by factoring the simplest terms first, such as pulling out common factors from each term.
• **Apply formulas**: Use formulas like \(a^2 - b^2 = (a - b)(a + b)\) to factor more complex polynomials.
• **Verification**: After factoring, always expand to check your work and ensure the result equals the original polynomial.

In the example problem, using the difference of squares formula, we factored \(81y^4 - 100z^2\) into \((9y^2 - 10z)(9y^2 + 10z)\). This step makes it easier to analyze the expression and solve related equations.