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Chapter 6: Problem 66

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

Short Answer

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

Step by step solution

01

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 \(b^2 = 100z^2\).
02

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 \((10z)^2 = 100z^2\).
03

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 \( (9y^2 - 10z)(9y^2 + 10z) \).
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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