Simplify Rational Expressions & Find Restrictions

Hey guys! Today, we're diving deep into the awesome world of rational expressions. We've got a cool problem that's going to test your skills in multiplying these beasts and, just as importantly, figuring out where they're undefined. So, let's get right into it!

Our mission, should we choose to accept it, is to multiply the following expression and identify all the restrictions on x:

$$\frac{2 x^2+5 x-3}{4 x^2-12 x+5} \cdot \frac{6 x^2-7 x-20}{3 x^2+13 x+12}$$

This might look a bit intimidating at first glance, with all those quadratic terms, but don't sweat it! The key to tackling these problems is to break them down into smaller, manageable parts. That means factoring each and every quadratic expression involved. Once we've got everything factored, multiplying becomes a piece of cake, and finding those restrictions will be super straightforward. Remember, restrictions happen when the denominator of any fraction in the original expression equals zero. So, we need to keep a close eye on those denominators throughout the process. Let's roll up our sleeves and get factoring!

Factoring the Numerators and Denominators

Alright, team, let's start by dissecting each polynomial. We'll tackle them one by one, making sure we get them factorized perfectly. This is the crucial first step, and getting it right will make the rest of the problem flow smoothly. For each quadratic expression $ax^2 + bx + c$, we're looking for two binomials that multiply together to give us back the original quadratic. We'll be using a mix of techniques, including the 'ac method' or simply by inspection, depending on which one feels quicker for each specific expression. Remember, the goal is to find two numbers that multiply to $ac$ and add up to $b$. Let's get into the nitty-gritty of each one:

1. Numerator 1: $2x^2 + 5x - 3$ We need two numbers that multiply to $(2)(-3) = -6$ and add up to $5$. Those numbers are $6$ and $-1$. So, we can rewrite the middle term: $2x^2 + 6x - x - 3$. Now, we factor by grouping: $2x(x + 3) - 1(x + 3) = \mathbf{(2x-1)(x+3)}$. Phew, that one's done! Keep this factored form handy. This is what will help us cancel out terms later.

2. Denominator 1: $4x^2 - 12x + 5$ Here, we need two numbers that multiply to $(4)(5) = 20$ and add up to $-12$. Those numbers are $-10$ and $-2$. Let's rewrite and group: $4x^2 - 10x - 2x + 5$. Factor by grouping: $2x(2x - 5) - 1(2x - 5) = \mathbf{(2x-1)(2x-5)}$. Another one bites the dust! Notice we already have a $(2x-1)$ factor here, just like in the first numerator. This is a good sign that things are lining up for cancellation. Keep your eyes peeled for these common factors; they're the key to simplification.

3. Numerator 2: $6x^2 - 7x - 20$ For this one, we need two numbers that multiply to $(6)(-20) = -120$ and add up to $-7$. This might take a little more thought. Let's list some factors of 120: (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12). Bingo! $8$ and $15$. Since we need a sum of $-7$, we'll use $-15$ and $8$. Rewrite and group: $6x^2 - 15x + 8x - 20$. Factor by grouping: $3x(2x - 5) + 4(2x - 5) = \mathbf{(3x+4)(2x-5)}$. Awesome! We found another $(2x-5)$ factor, matching the second denominator. This is exactly what we want to see for simplification. The more common factors we spot, the simpler this expression will become.

4. Denominator 2: $3x^2 + 13x + 12$ Lastly, we need two numbers that multiply to $(3)(12) = 36$ and add up to $13$. Let's think... factors of 36: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). Aha! $4$ and $9$ add up to $13$. So, we rewrite and group: $3x^2 + 9x + 4x + 12$. Factor by grouping: $3x(x + 3) + 4(x + 3) = \mathbf{(3x+4)(x+3)}$. We did it! All four quadratics are factored. We've got $(2x-1)(x+3)$ for the first numerator, $(2x-1)(2x-5)$ for the first denominator, $(3x+4)(2x-5)$ for the second numerator, and $(3x+4)(x+3)$ for the second denominator. Great job, everyone!

Multiplying the Factored Expressions

Now that we've got all our pieces factored, multiplying the two fractions is a breeze. We simply multiply the numerators together and the denominators together. Remember, we're not actually multiplying out the factors yet; we're keeping them in their factored form because that's how we'll spot the common terms to cancel. It's like having a bunch of LEGO bricks – you don't want to glue them together randomly; you want to see how they fit and build something cool.

So, our expression now looks like this:

$$\frac{(2x-1)(x+3)}{(2x-1)(2x-5)} \cdot \frac{(3x+4)(2x-5)}{(3x+4)(x+3)}$$

To multiply, we just combine everything into one big fraction:

$$\frac{(2x-1)(x+3)(3x+4)(2x-5)}{(2x-1)(2x-5)(3x+4)(x+3)}$$

Look at that! It's a beautiful display of factors, both upstairs and downstairs. This is where the magic of simplification happens. We can cancel out any factor that appears in both the numerator and the denominator. It's like dividing the same number by itself – it always equals 1!

Let's identify the common factors:

• $(2x-1)$ is in both the numerator and the denominator.
• $(x+3)$ is in both the numerator and the denominator.
• $(3x+4)$ is in both the numerator and the denominator.
• $(2x-5)$ is in both the numerator and the denominator.

Wait a minute... it looks like every single factor is common to both the top and the bottom! This is pretty neat. When all the factors cancel out like this, what are we left with? That's right, we're left with 1!

So, the simplified expression is:

$$\frac{\cancel{(2x-1)}\cancel{(x+3)}\cancel{(3x+4)}\cancel{(2x-5)}} {\cancel{(2x-1)}\cancel{(2x-5)}\cancel{(3x+4)}\cancel{(x+3)}} = 1$$

Boom! The multiplication and simplification result in a nice, clean $1$. Isn't math cool? But we're not done yet, guys. We still have one more critical step: finding the restrictions.

Determining the Restrictions on x

Now, for the super important part: identifying the restrictions on x. Remember, a rational expression is undefined whenever its denominator is equal to zero. This applies to any denominator that appears anywhere in the original expression before any simplification happens. It's like checking the ingredients list for allergies before you start cooking; you wouldn't want to find out later that something was off. So, we need to look back at the denominators of our original fractions and set each factor within them to zero.

Our original denominators were $4x^2 - 12x + 5$ and $3x^2 + 13x + 12$.

We already factored these, so let's use the factored forms:

1. From the first denominator, $4x^2 - 12x + 5$, we factored it into $(2x-1)(2x-5)$. Setting each factor to zero:

   • $2x - 1 = 0 ightarrow 2x = 1 ightarrow x = \frac{1}{2}$
   • $2x - 5 = 0 ightarrow 2x = 5 ightarrow x = \frac{5}{2}$

2. From the second denominator, $3x^2 + 13x + 12$, we factored it into $(3x+4)(x+3)$. Setting each factor to zero:

   • $3x + 4 = 0 ightarrow 3x = -4 ightarrow x = -\frac{4}{3}$
   • $x + 3 = 0 ightarrow x = -3$

So, the values of x that make the denominators zero are $\frac{1}{2}$, $\frac{5}{2}$, $-\frac{4}{3}$, and $-3$. These are the values that x cannot be equal to for the original expression to be defined.

It's crucial to note that even though the factors $(2x-1)$, $(2x-5)$, $(3x+4)$, and $(x+3)$ all cancelled out during simplification, they still create restrictions because they were part of the original denominators. The expression is undefined at these points, even if the simplified form looks like it's defined. Always go back to the original expression for restrictions, guys!

Final Answer and Conclusion

Putting it all together, the simplified expression is $1$, and the restrictions on x are $x \neq \frac{1}{2}, \frac{5}{2}, -\frac{4}{3}, -3$.

Let's look at the options provided:

A. $1$, $x \neq \frac{1}{2}, \frac{5}{2}, 3, -\frac{4}{3}$ B. $1$, $x \neq \frac{1}{2}, \frac{5}{2}, -\frac{4}{3}, -3$ C. $\frac{3 x+4}{2 x-1}$, $x \neqDiscussion category : mathematics

Comparing our findings with the options, option B matches exactly what we calculated. The simplified form is $1$, and the restrictions are indeed $x \neq \frac{1}{2}, \frac{5}{2}, -\frac{4}{3}, -3$.

This problem really highlights the importance of thorough factoring and understanding where restrictions come from. It's not just about simplifying; it's about understanding the behavior of the function across its entire domain. Keep practicing these, and you'll become rational expression masters in no time! Keep up the great work, and I'll catch you in the next one!