Simplifying Rational Expressions: A Step-by-Step Guide

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Simplifying Rational Expressions: A Comprehensive Guide

Hey math enthusiasts! Today, we're diving into the world of algebraic expressions, specifically focusing on how to divide them. We're going to break down the expression 2x2+5x+3x2βˆ’3xβˆ’4Γ·4x2+2xβˆ’6x2βˆ’8x+16\frac{2 x^2+5 x+3}{x^2-3 x-4} \div \frac{4 x^2+2 x-6}{x^2-8 x+16} step-by-step, making sure you understand every move. Don't worry, it might look a bit intimidating at first, but trust me, it's totally manageable. Just stick with me, and we'll conquer this together! This process involves factoring, simplifying, and remembering a few key rules. So, grab your notebooks, and let's get started. We'll be using factoring techniques, and the rules of fraction division. By the end of this guide, you'll be comfortable with dividing rational expressions and simplifying them. So, let’s get into it, shall we?

Step 1: Understanding the Problem and the Strategy

Our goal is to divide the given rational expressions. The expression involves polynomials in both the numerator and the denominator. The first key to solving this type of problem is to understand that dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of a fraction is simply flipping the numerator and the denominator. So, the first move is to rewrite the division problem as a multiplication problem, flipping the second fraction (the divisor). The original expression is: 2x2+5x+3x2βˆ’3xβˆ’4Γ·4x2+2xβˆ’6x2βˆ’8x+16\frac{2 x^2+5 x+3}{x^2-3 x-4} \div \frac{4 x^2+2 x-6}{x^2-8 x+16}. We rewrite this by changing the division to multiplication and inverting the second fraction. This gives us 2x2+5x+3x2βˆ’3xβˆ’4Γ—x2βˆ’8x+164x2+2xβˆ’6\frac{2 x^2+5 x+3}{x^2-3 x-4} \times \frac{x^2-8 x+16}{4 x^2+2 x-6}. Notice how the division sign has changed to multiplication, and the second fraction has been 'flipped.' This is the cornerstone of dividing rational expressions: convert the division into multiplication by the reciprocal. This simple change allows us to apply the familiar rules of multiplying fractions: multiply the numerators and multiply the denominators. Once this is done, the next major step involves factoring the polynomial expressions. Factoring helps us find common factors that can be cancelled out, simplifying the expression to its lowest terms. So the main strategy involves three core steps: rewrite the division as multiplication by the reciprocal, factor all the polynomials, and simplify by cancelling out common factors. The success of this method hinges on your ability to factor polynomials accurately. Let’s get into the factoring part, shall we?

Step 2: Factorizing the Polynomials

Factoring is the heart of simplifying rational expressions. We need to factor each of the polynomials in the numerators and denominators. This means breaking down each polynomial into a product of simpler expressions. The key here is to identify and factor out any common factors, as well as use factoring techniques like the 'ac method' or recognizing special forms like difference of squares or perfect square trinomials. Let's start by factoring each part of our expression. We have four polynomials to factor: 2x2+5x+32x^2 + 5x + 3, x2βˆ’3xβˆ’4x^2 - 3x - 4, 4x2+2xβˆ’64x^2 + 2x - 6, and x2βˆ’8x+16x^2 - 8x + 16.

  • Factoring 2x2+5x+32x^2 + 5x + 3: We're looking for two binomials that multiply to give us this quadratic expression. Using the ac method (where you multiply the coefficient of x2x^2 by the constant term), we look for two numbers that multiply to 2βˆ—3=62*3=6 and add up to 5. Those numbers are 2 and 3. We rewrite the middle term and factor by grouping: 2x2+2x+3x+3=2x(x+1)+3(x+1)=(2x+3)(x+1)2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1).
  • Factoring x2βˆ’3xβˆ’4x^2 - 3x - 4: We look for two numbers that multiply to -4 and add to -3. Those numbers are -4 and 1. So, this factors into (xβˆ’4)(x+1)(x - 4)(x + 1).
  • Factoring 4x2+2xβˆ’64x^2 + 2x - 6: First, factor out the greatest common factor, which is 2: 2(2x2+xβˆ’3)2(2x^2 + x - 3). Now, we factor the quadratic inside the parentheses. We are looking for two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2. Thus, 2x2+xβˆ’32x^2 + x - 3 factors into (2x+3)(xβˆ’1)(2x + 3)(x - 1). Therefore, 4x2+2xβˆ’6=2(2x+3)(xβˆ’1)4x^2 + 2x - 6 = 2(2x + 3)(x - 1).
  • Factoring x2βˆ’8x+16x^2 - 8x + 16: This is a perfect square trinomial. It factors into (xβˆ’4)2(x - 4)^2 or (xβˆ’4)(xβˆ’4)(x - 4)(x - 4).

Now that we have factored all the polynomials, we are ready for the next step: substituting the factored forms back into our expression. Remember, accuracy in factoring is crucial! Make sure to double-check your factoring to avoid mistakes. Alright, are you ready to substitute the factored forms into the expressions?

Step 3: Substitute and Simplify the Expression

Now that we've factored all the polynomials, it's time to substitute these factored forms back into our expression. Recall that we rewrote the division as multiplication by the reciprocal, so we have 2x2+5x+3x2βˆ’3xβˆ’4Γ—x2βˆ’8x+164x2+2xβˆ’6\frac{2 x^2+5 x+3}{x^2-3 x-4} \times \frac{x^2-8 x+16}{4 x^2+2 x-6}. After factoring we found that this equals (2x+3)(x+1)(xβˆ’4)(x+1)Γ—(xβˆ’4)(xβˆ’4)2(2x+3)(xβˆ’1)\frac{(2x + 3)(x + 1)}{(x - 4)(x + 1)} \times \frac{(x - 4)(x - 4)}{2(2x + 3)(x - 1)}. Now we have the expression with each part factored. The next step is to simplify by canceling out any common factors between the numerators and the denominators. This is where the magic happens! We're looking for identical expressions in both the numerator and the denominator that we can 'cancel' or divide out, as they effectively equal 1. So, let’s go through the expression step-by-step. Let's see what we can cancel: (2x+3)(2x + 3) appears in the numerator and the denominator, so we can cancel them out. (x+1)(x + 1) also appears in both the numerator and denominator, so we can cancel it out. And finally, (xβˆ’4)(x - 4) appears in the numerator and denominator. We can also cancel this out. After canceling out common factors, our expression now becomes 11Γ—(xβˆ’4)2(xβˆ’1)\frac{1}{1} \times \frac{(x - 4)}{2(x - 1)}. Which means that after canceling the common factors, we are left with xβˆ’42(xβˆ’1)\frac{x-4}{2(x-1)}. Remember that you can only cancel out factors, not terms that are added or subtracted. The simplified expression is the final answer! Now, what do you think? It's not that hard, right? Make sure you check for any remaining opportunities to simplify before moving on. Did you follow all the steps? If you did, then you are ready to complete the last step.

Step 4: Final Answer and Conclusion

We started with a complex rational expression and, through a series of logical steps, we've simplified it significantly. The original expression 2x2+5x+3x2βˆ’3xβˆ’4Γ·4x2+2xβˆ’6x2βˆ’8x+16\frac{2 x^2+5 x+3}{x^2-3 x-4} \div \frac{4 x^2+2 x-6}{x^2-8 x+16} has been simplified to xβˆ’42(xβˆ’1)\frac{x-4}{2(x-1)}. This is the final simplified form of the expression. So, the key takeaways from this exercise are: understand the rules of dividing fractions and how to turn a division problem into a multiplication problem, factor each polynomial expression to identify common factors, and cancel those common factors to simplify the expression. Remember, practice is key! The more you practice, the more comfortable and confident you'll become in solving these types of problems. Go back and revisit the steps to ensure you completely understand each part. Don't be afraid to try more examples! The goal is to build your confidence and become a pro at simplifying rational expressions. Keep practicing, and you will get there! You can always go back and review the steps we took today, so you don't forget the process. Keep up the excellent work, and always remember: Mathematics is a journey, not a destination. So have fun with it, guys!