Simplifying Rational Expressions A Step By Step Guide

Hey guys! Today, we're diving into the world of rational expressions, specifically focusing on simplifying quotients and figuring out when these expressions actually exist. We'll break down a problem step-by-step, making sure everyone can follow along. So, let's jump right in!

Understanding the Problem

We're going to tackle the following expression:

$\frac{x^2-2 x-8}{x^2-2 x-15} \div \frac{2 x^2-8 x}{2 x^2-10 x}$

Our main goals are to:

1. Simplify this quotient as much as possible.
2. Identify the numerator and denominator of the simplified form.
3. Determine the values of x for which the expression is undefined.

Sounds like a plan? Great! Let's get started with simplifying.

Step 1: Rewriting Division as Multiplication

Remember the golden rule of dividing fractions? We flip the second fraction and multiply! This is the same for rational expressions. So, we rewrite our expression as:

$\frac{x^2-2 x-8}{x^2-2 x-15} \times \frac{2 x^2-10 x}{2 x^2-8 x}$

This simple change makes the whole process a lot easier. Now, we're dealing with multiplication, which is more straightforward to handle. Next up, we'll factor everything in sight!

Step 2: Factoring Quadratics and Expressions

Factoring is the heart of simplifying rational expressions. We need to break down each quadratic and expression into its simplest factors. This will allow us to cancel out common terms later on. Let's take it piece by piece:

Factoring the First Numerator: $x^2 - 2x - 8$

We need two numbers that multiply to -8 and add up to -2. Those numbers are -4 and 2. So, we can factor this as:

$x^2 - 2x - 8 = (x - 4)(x + 2)$

Factoring the First Denominator: $x^2 - 2x - 15$

Similarly, we need two numbers that multiply to -15 and add up to -2. These are -5 and 3. Thus, we factor this as:

$x^2 - 2x - 15 = (x - 5)(x + 3)$

Factoring the Second Numerator: $2x^2 - 10x$

Here, we can first factor out a common factor of 2x:

$2x^2 - 10x = 2x(x - 5)$

Factoring the Second Denominator: $2x^2 - 8x$

Again, we factor out the common factor, which is 2x:

$2x^2 - 8x = 2x(x - 4)$

Now that we've factored everything, our expression looks like this:

$\frac{(x - 4)(x + 2)}{(x - 5)(x + 3)} \times \frac{2x(x - 5)}{2x(x - 4)}$

See how much more manageable it looks? Time for the fun part – canceling!

Step 3: Canceling Common Factors

This is where the magic happens! We look for factors that appear in both the numerator and the denominator and cancel them out. Remember, we can only cancel factors that are multiplied, not added or subtracted. Let's identify the common factors:

• (x - 4) appears in the first numerator and the second denominator.
• (x - 5) appears in the first denominator and the second numerator.
• 2x appears in both the second numerator and the second denominator.

Canceling these out, we get:

$\frac{\cancel{(x - 4)}(x + 2)}{\cancel{(x - 5)}(x + 3)} \times \frac{\cancel{2x}\cancel{(x - 5)}}{\cancel{2x}\cancel{(x - 4)}} = \frac{x + 2}{x + 3}$

So, our simplified expression is $\frac{x + 2}{x + 3}$.

We've nailed the first part! The simplified form has a numerator of (x + 2) and a denominator of (x + 3). Now, let's figure out when this expression exists.

Step 4: Determining When the Expression Exists

A rational expression exists as long as its denominator is not zero. We can't divide by zero, it's a big no-no in math! So, we need to find the values of x that would make the original denominators zero. This includes the denominators before we simplified the expression.

Let's go back to our factored form before canceling:

$\frac{(x - 4)(x + 2)}{(x - 5)(x + 3)} \times \frac{2x(x - 5)}{2x(x - 4)}$

We have the following denominators to consider:

1. $x^2 - 2x - 15 = (x - 5)(x + 3)$
2. $2x^2 - 8x = 2x(x - 4)$
3. $2x^2 - 10x = 2x(x-5)$

Let's find when each of these equals zero.

Denominator 1: $(x - 5)(x + 3) = 0$

This gives us two solutions:

• x - 5 = 0 => x = 5
• x + 3 = 0 => x = -3

Denominator 2: $2x(x - 4) = 0$

This gives us two solutions:

• 2x = 0 => x = 0
• x - 4 = 0 => x = 4

Denominator 3: $2x(x-5) = 0$

This gives us two solutions:

• 2x = 0 => x = 0
• x - 5 = 0 => x = 5

So, the values of x that make the denominators zero are: -3, 0, 4, and 5. Therefore, the expression does not exist when x = -3, 0, 4, or 5.

Final Answers

Let's wrap it all up:

• The simplified form of the quotient has a numerator of (x + 2) and a denominator of (x + 3).
• The expression does not exist when x = -3, 0, 4, 5.

Conclusion

We've successfully simplified a rational expression and identified the values for which it's undefined. Remember, the key steps are:

1. Rewrite division as multiplication by flipping the second fraction.
2. Factor all numerators and denominators completely.
3. Cancel out common factors.
4. Find the values that make the original denominators zero.

This might seem like a lot, but with practice, you'll become a pro at simplifying rational expressions. Keep up the great work, guys!