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Chapter 7: Problem 77

Multiply or divide as indicated. $$ \left(z^{2}-1\right) \cdot \frac{1}{1-z} $$

Short Answer

Expert verified
-(z+1)

Step by step solution

01

Factor the expression

Factor the numerator of the given expression. The numerator is \(z^{2} - 1\), which is a difference of squares. It can be factored into \( (z-1)(z+1) \). So the expression becomes, \[ (z-1)(z+1) \frac{1}{1-z} \]
02

Simplify the denominator

Notice that \( 1-z \) can be rewritten as \ -(z-1) \. This is because \( z-1 \) changes sign when multiplied by \ -1 \. So, substituting \ -(z-1) \ in place of \( 1-z \), the expression now becomes, \[ (z-1)(z+1) \frac{1}{-(z-1)} \]
03

Simplify the expression

Simplify the expression by canceling out \( z-1 \) from the numerator and the denominator, which leaves us with, \[ \frac{(z+1)}{ -1} \]
04

Final simplification

Finally, divide \( z+1 \) by \ -1 \ to get the simplified result, \[ -(z+1) \]

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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 an important method for factoring. It occurs when you have an expression in the form of \( a^2 - b^2 \). To factor such expressions, you can use the formula \( a^2 - b^2 = (a - b)(a + b) \). Let's look at the given problem. The original expression is \( z^2 - 1 \). Here, \( a \) is \( z \) and \( b \) is 1, so we have \( z^2 - 1^2 \). Applying the difference of squares, we factor it as \( (z - 1)(z + 1) \).
factoring
Factoring is the process of breaking down an expression into products of simpler expressions. In our exercise, the term \( z^2 - 1 \) was factored into \( (z - 1)(z + 1) \) using the difference of squares method. Factoring is useful because it can simplify complex algebraic expressions, making them easier to work with. Always look for common patterns like difference of squares, perfect square trinomials, or common factors when factoring.
simplifying algebraic fractions
Simplifying algebraic fractions involves reducing them to their simplest form. Our example includes the term \( \frac{1}{1-z} \). Notice that \( 1 - z \) can be rewritten as \( -(z - 1) \) because of the property that switching the terms and factoring out a negative sign gives us the same value. This helps to cancel out the common factors in the numerator and the denominator. For the given problem, we simplified \( (z-1)(z+1) \frac{1}{-(z-1)} \) to \( \frac{z+1}{-1} \) by canceling out the \( z-1 \) terms.
expression simplification
Expression simplification involves rewriting an expression in a simpler or more standard form. In the final steps of our problem, we took \( \frac{z+1}{-1} \) and applied a simple division to get \( -(z+1) \). Simplification often involves combining like terms, canceling out common factors, or using algebraic identities. The goal is to make the expression easier to work with in future calculations or to provide a clear, final result that's easy to understand.

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Most popular questions from this chapter

For \(k>0,\) if \(y\) varies inversely as \(x,\) then when \(x\) increases, \(y\) _____ and when \(x\) decreases, \(y\) _____.

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