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Unit 4
Lesson 6: Multiplying and dividing rational expressions
Math>Precalculus>Rational functions>Multiplying and dividing rational expressions
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Multiply & divide rational expressions
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Problem
Answer two questions about the following rational division.
 
 
 
\[\dfrac{\dfrac{x+8}{x^2-6x-7}}{\,\,\,\dfrac{x^2+16x+64}{x+1}\,\,\,}\]
1. What is the quotient in lowest terms?
2. What values of 
\[x\] must we exclude from the domains of the expressions?
Choose all answers that apply:
Choose all answers that apply:
(Choice A)   
\[x=-8\]
A
\[x=-8\]
(Choice B)   
\[x=-1\]
B
\[x=-1\]
(Choice C)   
\[x=1\]
C
\[x=1\]
(Choice D)   
\[x=7\]
D
\[x=7\]
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Do 4 problems
You completed your assignment for PD3 - IM3.

Let's tackle the given rational expression step by step.

The problem involves dividing two rational expressions:

\[
\dfrac{\dfrac{x+8}{x^2-6x-7}}{\dfrac{x^2+16x+64}{x+1}}
\]

### Step 1: Simplify the expression
To divide by a fraction, we multiply by the reciprocal of the second fraction:

\[
\dfrac{x+8}{x^2-6x-7} \times \dfrac{x+1}{x^2+16x+64}
\]

Now, let's factor the quadratic expressions:

1. **Factor \( x^2 - 6x - 7 \):**

   We need two numbers that multiply to \( -7 \) and add up to \( -6 \). These numbers are \( -7 \) and \( 1 \). So, we can factor it as:

   \[
   x^2 - 6x - 7 = (x - 7)(x + 1)
   \]

2. **Factor \( x^2 + 16x + 64 \):**

   This is a perfect square trinomial, so it factors as:

   \[
   x^2 + 16x + 64 = (x + 8)^2
   \]

Thus, the expression becomes:

\[
\dfrac{x+8}{(x-7)(x+1)} \times \dfrac{x+1}{(x+8)^2}
\]

### Step 2: Cancel out common factors
We can cancel out the terms that appear in both the numerator and denominator:

- \( (x + 8) \) appears in the numerator of the first fraction and the denominator of the second fraction.
- \( (x + 1) \) appears in the denominator of the first fraction and the numerator of the second fraction.

After canceling, we are left with:

\[
\dfrac{1}{(x-7)(x+8)}
\]

This is the quotient in its lowest terms.

### Step 3: Determine the excluded values
We must exclude values of \( x \) that make any denominator equal to zero in the original expression.

- From \( x^2 - 6x - 7 = (x-7)(x+1) \), we get the potential values \( x = 7 \) and \( x = -1 \).
- From \( x^2 + 16x + 64 = (x+8)^2 \), we get the potential value \( x = -8 \).

Therefore, the values of \( x \) that must be excluded from the domain are \( x = -8 \), \( x = -1 \), and \( x = 7 \).

### Final Answer:
1. The quotient in lowest terms is:

   \[
   \dfrac{1}{(x - 7)(x + 8)}
   \]

2. The values of \( x \) that must be excluded are:
   - **Choice A:** \( x = -8 \)
   - **Choice B:** \( x = -1 \)
   - **Choice D:** \( x = 7 \)

Would you like further details on how to approach similar problems or additional explanations? 

### Related questions:
1. How do you simplify complex rational expressions involving multiple terms?
2. What is the domain of a rational expression, and how do we find it?
3. How can factoring help in simplifying rational expressions?
4. What is the process of multiplying rational expressions, and how is it different from division?
5. How do we identify and exclude values from the domain in rational functions?

**Tip:** Always check for factors that can be canceled out before simplifying to make your work easier.

Answer two questions about the following rational division. 

\[\dfrac{\dfrac{x+8}{x^2-6x-7}}{\dfrac{x^2+16x+64}{x+1}}\]
1. What is the quotient in lowest terms? 2. What values of \[x\] must we exclude from the domains of the expressions?

This page explains how to simplify a rational expression involving division and factoring. Learn to identify excluded values from the domain of rational expressions, and see the step-by-step process for finding the quotient in its lowest terms. Suitable for students in Grades 9-12.

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