Simplifying Rational Expressions Find Numerator And Denominator

Hey guys! Let's dive into simplifying a rational expression problem that often pops up in algebra. We'll break it down step-by-step, so it's super clear. Our mission is to simplify the following quotient:

$\frac{3 x^2-27 x}{2 x^2+13 x-7} \div \frac{3 x}{4 x^2-1}$

We'll tackle this by factoring, simplifying, and identifying any restrictions on the variable x. Let's get started!

1. Factoring the Numerator and Denominator

The first key step in simplifying rational expressions is factoring. Factoring helps us break down complex polynomials into simpler terms that we can then cancel out. Let’s look at each part of our expression individually.

Factoring the First Numerator: $3x^2 - 27x$

When you see an expression like $3x^2 - 27x$, always look for a common factor first. In this case, both terms have a common factor of $3x$. Factoring this out, we get:

$3x^2 - 27x = 3x(x - 9)$

This is much simpler already! We've turned a quadratic expression into a product of a monomial and a binomial. Remember, factoring is like reverse distribution, so you can always check your work by distributing the $3x$ back into $(x - 9)$ to make sure you get the original expression.

Factoring the First Denominator: $2x^2 + 13x - 7$

Now, let's tackle the quadratic expression $2x^2 + 13x - 7$. This one is a bit trickier because the leading coefficient (the number in front of $x^2$) is not 1. We need to find two numbers that multiply to the product of the leading coefficient and the constant term (2 * -7 = -14) and add up to the middle coefficient (13). Those numbers are 14 and -1.

So, we rewrite the middle term using these numbers:

$2x^2 + 13x - 7 = 2x^2 + 14x - x - 7$

Next, we factor by grouping. We group the first two terms and the last two terms:

$(2x^2 + 14x) + (-x - 7)$

Factor out the greatest common factor (GCF) from each group:

$2x(x + 7) - 1(x + 7)$

Notice that both terms now have a common factor of $(x + 7)$. We factor this out:

$(2x - 1)(x + 7)$

So, the factored form of $2x^2 + 13x - 7$ is $(2x - 1)(x + 7)$.

Factoring the Second Numerator: $3x$

The numerator $3x$ is already in its simplest form. There's nothing more to factor here!

Factoring the Second Denominator: $4x^2 - 1$

This expression, $4x^2 - 1$, is a classic example of the difference of squares. The difference of squares pattern is $a^2 - b^2 = (a - b)(a + b)$. In our case, $4x^2$ is $(2x)^2$ and 1 is $1^2$. Applying the pattern, we get:

$4x^2 - 1 = (2x - 1)(2x + 1)$

Now we have factored all parts of the original expression. Let's put it all together.

2. Rewriting the Division as Multiplication

Before we simplify, we need to deal with the division. Dividing by a fraction is the same as multiplying by its reciprocal. So, we rewrite the original expression as a multiplication:

$\frac{3 x^2-27 x}{2 x^2+13 x-7} \div \frac{3 x}{4 x^2-1} = \frac{3 x^2-27 x}{2 x^2+13 x-7} \times \frac{4 x^2-1}{3 x}$

Now, substitute the factored forms we found earlier:

$\frac{3x(x - 9)}{(2x - 1)(x + 7)} \times \frac{(2x - 1)(2x + 1)}{3x}$

This looks like a big fraction, but now we're set up to cancel out common factors!

3. Simplifying the Expression

Now comes the fun part: canceling out common factors. We have the expression:

$\frac{3x(x - 9)}{(2x - 1)(x + 7)} \times \frac{(2x - 1)(2x + 1)}{3x}$

Look for factors that appear in both the numerator and the denominator. We can cancel out:

• $3x$ in the numerator and denominator
• $(2x - 1)$ in the numerator and denominator

After canceling, we're left with:

$\frac{(x - 9)}{(x + 7)} \times \frac{(2x + 1)}{1}$

Multiply the remaining terms:

$\frac{(x - 9)(2x + 1)}{(x + 7)}$

So, the simplified form of the quotient is $\frac{(x - 9)(2x + 1)}{(x + 7)}$.

4. Identifying the Numerator and Denominator

The question asks for the simplest form of the quotient and wants us to identify the numerator and denominator separately. From our simplified expression, $\frac{(x - 9)(2x + 1)}{(x + 7)}$, we can clearly see:

• Numerator: $(x - 9)(2x + 1)$
• Denominator: $(x + 7)$

We can expand the numerator if desired, but leaving it in factored form often makes it easier to analyze the expression. Expanding the numerator, we get:

$(x - 9)(2x + 1) = 2x^2 + x - 18x - 9 = 2x^2 - 17x - 9$

So, the numerator can also be written as $2x^2 - 17x - 9$.

5. Restrictions on the Variable

Before we declare victory, we need to consider any restrictions on the variable x. Restrictions occur when the denominator of the original expression (or any intermediate step before simplification) equals zero. This would make the expression undefined because division by zero is a big no-no in math.

Looking back at the original expression and our factored forms, we had denominators of:

• $2x^2 + 13x - 7 = (2x - 1)(x + 7)$
• $3x$
• $4x^2 - 1 = (2x - 1)(2x + 1)$

We need to find the values of x that make these denominators zero. Setting each factor to zero gives us:

1. $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
2. $x + 7 = 0 \Rightarrow x = -7$
3. $3x = 0 \Rightarrow x = 0$
4. $2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$

So, the values of x that make the denominators zero are $x = \frac{1}{2}$, $x = -7$, $x = 0$, and $x = -\frac{1}{2}$. These are the values that x cannot be.

Therefore, the restrictions on x are:

$x \neq \frac{1}{2}, -7, 0, -\frac{1}{2}$

Conclusion

To wrap things up, the simplest form of the given quotient has:

• Numerator: $(x - 9)(2x + 1)$ or $2x^2 - 17x - 9$
• Denominator: $(x + 7)$
• Restrictions on x: $x \neq \frac{1}{2}, -7, 0, -\frac{1}{2}$

Simplifying rational expressions involves factoring, rewriting division as multiplication, canceling common factors, and identifying restrictions. It might seem like a lot of steps, but with practice, it becomes second nature. Keep up the great work, and you'll master these problems in no time!

Remember, understanding the step-by-step process is crucial for tackling more complex problems. Factoring is your best friend in simplifying rational expressions, so make sure you're comfortable with different factoring techniques. Also, don't forget to consider restrictions on the variable to ensure your solution is complete and accurate.

Happy simplifying, and keep crushing those math problems!