Solving Rational Expressions: A Step-by-Step Guide

by Andrew McMorgan 51 views

Hey guys! Today, we're diving into the fascinating world of rational expressions. Specifically, we're going to tackle a problem that involves dividing two rational expressions. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step, so you can follow along and master this skill. Our example problem is: 12x+127x2βˆ’23xβˆ’20Γ·3x2βˆ’27xβˆ’307x+5\frac{12 x+12}{7 x^2-23 x-20} \div \frac{3 x^2-27 x-30}{7 x+5}. Let's get started!

Understanding Rational Expressions

Before we jump into the solution, let's make sure we're all on the same page about what rational expressions are. A rational expression is simply a fraction where the numerator and denominator are polynomials. Think of it like regular fractions, but with variables and exponents involved. For example, x+1xβˆ’2\frac{x+1}{x-2} and 3x2+2xβˆ’1x2+4\frac{3x^2 + 2x - 1}{x^2 + 4} are both rational expressions. When we're dealing with operations like division, we need to remember some key principles, just like with regular fractions. The most important thing to remember when dividing fractions is that you multiply by the reciprocal of the second fraction. This same rule applies to rational expressions. We will also use factoring, which is the process of breaking down a polynomial into its constituent factors. Factoring makes it easier to simplify rational expressions and to identify common terms that can be canceled out. Factoring is a critical step because it allows us to simplify the expression and make it easier to work with. Without factoring, we'd be stuck with complex polynomials that are difficult to manipulate. Understanding the fundamentals of rational expressions and factoring is crucial for solving these types of problems effectively. This sets the stage for a smooth and accurate solution. Are you ready to take the plunge and unravel the solution to our main problem? Let’s do it!

Step 1: Rewriting Division as Multiplication

The first crucial step in tackling this problem is to remember our golden rule for dividing fractions: division is the same as multiplying by the reciprocal. So, instead of dividing by 3x2βˆ’27xβˆ’307x+5\frac{3 x^2-27 x-30}{7 x+5}, we're going to multiply by its reciprocal, which is 7x+53x2βˆ’27xβˆ’30\frac{7 x+5}{3 x^2-27 x-30}. This simple switch transforms our problem into something much more manageable. Our expression now looks like this: 12x+127x2βˆ’23xβˆ’20Γ—7x+53x2βˆ’27xβˆ’30\frac{12 x+12}{7 x^2-23 x-20} \times \frac{7 x+5}{3 x^2-27 x-30}. By changing the division to multiplication, we've set the stage for simplifying the expression. The next steps will involve factoring the polynomials in the numerators and denominators. Factoring helps us break down the expressions into their simplest components, making it easier to identify common factors that can be canceled out. Remember, the goal is to simplify the expression as much as possible, and this transformation is the first big step in that direction. We've effectively turned a complex division problem into a more straightforward multiplication problem, making the subsequent steps more accessible. With this transformation, we're ready to move on to the next stage: factoring the polynomials.

Step 2: Factoring the Polynomials

Now comes the fun part: factoring! We need to factor each polynomial in our expression to see if we can simplify things. Let's start with the first numerator, 12x+1212x + 12. We can factor out a 12, giving us 12(x+1)12(x + 1). Next up is the first denominator, 7x2βˆ’23xβˆ’207x^2 - 23x - 20. This is a quadratic expression, and we need to find two numbers that multiply to 7Γ—βˆ’20=βˆ’1407 \times -20 = -140 and add up to -23. After a little thought, we find that -28 and 5 fit the bill. So, we can rewrite the quadratic as 7x2βˆ’28x+5xβˆ’207x^2 - 28x + 5x - 20. Now, we factor by grouping: 7x(xβˆ’4)+5(xβˆ’4)7x(x - 4) + 5(x - 4), which simplifies to (7x+5)(xβˆ’4)(7x + 5)(x - 4). Moving on to the second numerator, 7x+57x + 5, there's nothing to factor hereβ€”it's already in its simplest form. Finally, let's tackle the second denominator, 3x2βˆ’27xβˆ’303x^2 - 27x - 30. We can start by factoring out a 3: 3(x2βˆ’9xβˆ’10)3(x^2 - 9x - 10). Now we need to factor the quadratic x2βˆ’9xβˆ’10x^2 - 9x - 10. We're looking for two numbers that multiply to -10 and add up to -9. Those numbers are -10 and 1. So, the factored form is 3(xβˆ’10)(x+1)3(x - 10)(x + 1). After factoring all the polynomials, our expression now looks like this:

12(x+1)(7x+5)(xβˆ’4)Γ—7x+53(xβˆ’10)(x+1)\frac{12(x + 1)}{(7x + 5)(x - 4)} \times \frac{7x + 5}{3(x - 10)(x + 1)}

Factoring is a critical step because it allows us to simplify the expression and to cancel out common factors. Without factoring, we'd be stuck with complex polynomials that are difficult to manipulate. This step transforms the expression into a format where we can easily identify terms that can be reduced, paving the way for a simpler solution. Are you ready to see how all this factoring pays off? Let's move on to the next step!

Step 3: Simplifying the Expression

Now for the satisfying part: simplifying! We've factored everything, so we can now cancel out common factors from the numerator and the denominator. Looking at our expression:

12(x+1)(7x+5)(xβˆ’4)Γ—7x+53(xβˆ’10)(x+1)\frac{12(x + 1)}{(7x + 5)(x - 4)} \times \frac{7x + 5}{3(x - 10)(x + 1)}

We can see that (x+1)(x + 1) appears in both the numerator and the denominator, so we can cancel them out. Also, (7x+5)(7x + 5) is present in both, so we can cancel that out too! Additionally, we can simplify the constants: 12 in the numerator and 3 in the denominator. 12 divided by 3 is 4, so we're left with 4 in the numerator. After canceling out these common factors, our expression simplifies to:

4(xβˆ’4)(xβˆ’10)\frac{4}{(x - 4)(x - 10)}

Wow, that's much simpler, isn't it? This step demonstrates the power of factoring. By breaking down the polynomials into their constituent parts, we were able to identify and eliminate common terms, leading to a significantly simplified expression. Simplifying is crucial because it reduces the complexity of the expression, making it easier to understand and work with. It's like decluttering a roomβ€”once you remove the unnecessary items, you can see the space more clearly. In this case, simplifying the rational expression allows us to see the core components more clearly, leading us to the final, most reduced form. Now that we've simplified our expression, we're just one step away from the final answer. Are you excited to see the result? Let’s wrap things up!

Step 4: The Final Answer

After all our hard work, we've arrived at the final answer! We started with a complex division problem, factored all the polynomials, canceled out common factors, and simplified the expression. The final simplified form is:

4(xβˆ’4)(xβˆ’10)\frac{4}{(x - 4)(x - 10)}

So, the correct answer is C) 4(xβˆ’4)(xβˆ’10)\frac{4}{(x-4)(x-10)}. Congratulations, guys! We did it! This problem demonstrates the importance of understanding rational expressions, factoring, and simplifying. Each step built upon the previous one, leading us to the final solution. By breaking down the problem into smaller, more manageable parts, we were able to tackle it effectively. This approach is valuable not only in mathematics but also in many other areas of life. Whether you're solving equations, writing code, or even planning a project, breaking it down into smaller steps can make the task much less daunting. Remember, practice makes perfect! The more you work with rational expressions, the more comfortable you'll become with factoring and simplifying. So, keep practicing, keep learning, and keep challenging yourselves. You've got this! Thanks for joining me on this mathematical adventure. I hope you found this explanation helpful and that you're now more confident in your ability to solve rational expressions. Until next time, keep exploring the wonderful world of math!