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Chapter 0: Problem 60

Factor the expression completely. $$ \left(a^{2}-1\right) b^{2}-4\left(a^{2}-1\right) $$

Short Answer

Expert verified
\((a - 1)(a + 1)(b - 2)(b + 2)\)

Step by step solution

01

Identify the Common Factor

The expression given is \( (a^2 - 1)b^2 - 4(a^2 - 1) \). Notice that \( (a^2 - 1) \) is a common factor in both terms.
02

Factor Out the Common Factor

Factor \( (a^2 - 1) \) out of the expression. The expression becomes: \[(a^2 - 1)(b^2 - 4)\]
03

Factor the Difference of Squares

Recognize that \( a^2 - 1 \) and \( b^2 - 4 \) are both differences of squares. \( a^2 - 1 \) can be factored as \( (a - 1)(a + 1) \). \( b^2 - 4 \) can be factored as \( (b - 2)(b + 2) \).
04

Write the Completely Factored Form

Combine these factors to write the expression in its completely factored form: \[(a - 1)(a + 1)(b - 2)(b + 2)\]

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Key Concepts

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

Common Factor
When factoring expressions, identifying a common factor is a great first step. A common factor is a term or expression that is consistent across the terms of an equation. It helps simplify expressions quickly.

In \[(a^2 - 1)b^2 - 4(a^2 - 1)\]identify \(a^2 - 1\); it appears in both terms. This means \(a^2 - 1\)is the common factor. By factoring it out, you reduce complexity and inch closer to solving the equation."
  • Determine recurring terms in all parts of the expression.
  • Extract and factor out these common terms.
Applying this concept simplifies the original expression to:\[(a^2 - 1)(b^2 - 4)\]This step allows you to easily focus on further factoring, as it reduces the expression into more manageable parts.
Difference of Squares
The concept of the "difference of squares" is a powerful technique in algebra. The difference of squares formula is: \[ a^2 - b^2 = (a-b)(a+b)\]This applies to expressions where two squared terms are subtracted from each other. It's a straightforward pattern but very useful.

In the given expression, you identified \(a^2 - 1\) and \(b^2 - 4\) as differences of squares. Here's how to factor them:
  • \(a^2 - 1\) recasts to \((a - 1)(a + 1)\) using \(1^2\) .
  • \(b^2 - 4\)turns into\((b - 2)(b + 2)\) because \(2^2 = 4\).
Each created factor represents a binomial, simplifying expressions considerably. This step greatly aids solving, as it allows you to express differences of squares in a completely factored form.
Completely Factored Form
Achieving the completely factored form requires combining all factored components into a single expression. After addressing both the common factor and difference of squares, you'll be left with an expression totally broken down.

In this exercise, the steps formed:
  • From\( (a^2 - 1)(b^2 - 4) \),
  • you break it further to \((a - 1)(a + 1)\)
  • and \((b - 2)(b + 2)\)
Now bring all components back together:\[(a - 1)(a + 1)(b - 2)(b + 2)\]
Completely factored expressions are paramount to solving and simplifying equations. They reveal roots or possible solutions and provide the simplest form of expressions. This efficient form is invaluable for deeper algebraic problem-solving.

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Most popular questions from this chapter

Simplify the compound fractional expression. $$ \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}} $$

Simplify the expression. (This type of expression arises in calculus when using the 鈥渜uotient rule.鈥�) $$ \frac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}} $$

Perform the multiplication or division and simplify. $$ \frac{x^{2}-2 x-15}{x^{2}-9} \cdot \frac{x+3}{x-5} $$

Perform the multiplication or division and simplify. $$ \frac{4 x}{x^{2}-4} \cdot \frac{x+2}{16 x} $$

Perform the addition or subtraction and simplify. $$ \frac{x}{x-4}-\frac{3}{x+6} $$
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