Problem 77
Question
For exercises 39-82, simplify. $$ \frac{z^{2}+6 z-16}{z-2} \div \frac{z+8}{z+3} $$
Step-by-Step Solution
Verified Answer
The simplified expression is \( z + 3 \).
1Step 1: Simplify the Division
To simplify the given expression, first rewrite the division of fractions as multiplication by the reciprocal. The given expression is \[ \frac{z^{2}+6z-16}{z-2} \times \frac{z+3}{z+8} \]
2Step 2: 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 \[ \frac{(z+8)(z-2)}{z-2} \times \frac{z+3}{z+8} \]
3Step 3: 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 \[ \frac{z+3}{1} = z+3 \]
Key Concepts
Factoring Quadratic ExpressionsDivision of FractionsCanceling Common Factors
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).
This step is crucial as it allows us to simplify expressions later on by revealing common factors that can be canceled.
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 \(8 + (-2) = 6\).
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 \(\frac{z^{2}+6z-16}{z-2} \/ \frac{1}{(z+8)/(z+3)}\).
By converting the division to multiplication, the problem becomes easier to handle, especially when factoring expressions later on.
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 \(\frac{z+3}{z+8}\).
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:
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.
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.
Other exercises in this chapter
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