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Chapter 7: Problem 77

For exercises 39-82, simplify. $$ \frac{z^{2}+6 z-16}{z-2} \div \frac{z+8}{z+3} $$

Short Answer

Expert verified
The simplified expression is \( z + 3 \).

Step by step solution

01

Simplify the Division

To simplify the given expression, first rewrite the division of fractions as multiplication by the reciprocal. The given expression is \[ \frac{z^{2}+6z-16}{z-2} \times \frac{z+3}{z+8} \]
02

Factorize the Quadratic Expression

Factor the quadratic expression in the numerator of the first fraction. We need to factor \(z^2 + 6z - 16\). This can be written as \[ (z+8)(z-2) \]. So the expression now is \[ \frac{(z+8)(z-2)}{z-2} \times \frac{z+3}{z+8} \]
03

Cancel Common Factors

Now, cancel out the common factors in the numerator and the denominator. The common factors \((z-2)\) and \((z+8)\) cancel each other out. The remaining expression is \[ \frac{z+3}{1} = z+3 \]

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Key Concepts

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

Factoring Quadratic Expressions
Factoring quadratic expressions is a key skill in algebra. It involves breaking down a polynomial such as \(z^2 + 6z - 16\) into simpler terms that, when multiplied together, give the original polynomial.

To factor \(z^2 + 6z - 16\), we look for two numbers that multiply to \(-16\) (the constant term) and add to \(6\) (the coefficient of the middle term).
  • The numbers 8 and -2 satisfy these conditions because \(8 \times -2 = -16\) and \(8 + (-2) = 6\).
Hence, \(z^2 + 6z - 16\) can be written as \((z + 8)(z - 2)\).

This step is crucial as it allows us to simplify expressions later on by revealing common factors that can be canceled.
Division of Fractions
When dividing fractions, the process becomes more straightforward if we convert the division into a multiplication by the reciprocal.

For instance, the expression \(\frac{z^{2}+6z-16}{z-2} \/ \frac{z+8}{z+3}\) can be rewritten as \(\frac{z^{2}+6z-16}{z-2} \/ \frac{1}{(z+8)/(z+3)}\).
  • The reciprocal of \(\frac{z+8}{z+3}\) is \(\frac{z+3}{z+8}\).
So, the original expression is now \(\frac{z^{2}+6z-16}{z-2} \times \frac{z+3}{z+8}\).

By converting the division to multiplication, the problem becomes easier to handle, especially when factoring expressions later on.
Canceling Common Factors
Canceling common factors simplifies expressions by removing identical terms from the numerator and the denominator.

Consider the factored expression \(\frac{(z + 8)(z - 2)}{z - 2} \times \frac{z + 3}{z + 8}\). Here, we see common factors:
  • The \((z - 2)\) in the numerator and denominator, which can be canceled out.
  • Similarly, \((z + 8)\) also appears in both the numerator and denominator and can be canceled.

After canceling, the simplified expression left is \(z + 3\).

Canceling common factors is an essential step in simplifying algebraic expressions. It reduces the complexity by removing redundant terms, making the expression easier to interpret and solve.

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Most popular questions from this chapter

Explain why the relationship of the number of square feet of carpet that need to be vacuumed, \(x\), and the amount of time it takes to vacuum the carpet, \(y\), is a direct variation.

When a car travels a fixed distance, the relationship between the speed of the car, \(x\), and the time it travels, \(y\), is an inverse variation. When the speed is \(\frac{48 \mathrm{mi}}{1 \mathrm{hr}}\), the time is \(0.75 \mathrm{hr}\). a. Find the constant of proportionality. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the time in hours to travel this distance at a speed of \(\frac{80 \mathrm{mi}}{1 \mathrm{hr}}\). d. Change the time in part \(\mathrm{c}\) to minutes.

For exercises \(35-36, T=\frac{336 \mathrm{gm}}{R}\) represents the relationship of tire diameter, \(T\); gear ratio, \(g\); speed, \(m\); and revolutions of the tire per minute, \(R\). Is the relationship of the given variables a direct variation or an inverse variation? $$ g \text { and } m \text { are constant; the relationship of } T \text { and } R $$

For exercises 79-82, (a) clear the fractions and solve. (b) check. $$ 1=\frac{7}{6} w+\frac{5}{12} $$

For exercises \(35-36, T=\frac{336 \mathrm{gm}}{R}\) represents the relationship of tire diameter, \(T\); gear ratio, \(g\); speed, \(m\); and revolutions of the tire per minute, \(R\). Is the relationship of the given variables a direct variation or an inverse variation? $$ g \text { and } R \text { are constant; the relationship of } T \text { and } m \text {. } $$
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