Using the Quotient Rule for Division

Oct 13, 2025 by thesweetdesigns.com 95 views

Introduction

When dealing with complex fractions, it's often necessary to simplify them to make them more manageable. In this case, we're given the expression $\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1}$ and we need to find an equivalent expression if no denominator equals zero. This involves using the rules of division and simplifying the resulting expression.

Understanding the Problem

To begin, let's break down the given expression and understand what's being asked. We have a complex fraction, which is a fraction that contains one or more fractions in its numerator or denominator. In this case, we have two fractions: $\frac{x+3}{x^2-2x-3}$ and $\frac{x^2+2x-3}{x+1}$ . We're asked to find an equivalent expression for the division of these two fractions.

Using the Quotient Rule for Division

To simplify the given expression, we can use the quotient rule for division, which states that division of two fractions is equivalent to multiplication by the reciprocal of the divisor. In other words, $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ . Applying this rule to the given expression, we get:

\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1} = \frac{x+3}{x^2-2x-3} \cdot \frac{x+1}{x^2+2x-3}

Simplifying the Expression

Now that we have the expression in a simpler form, we can simplify it further by multiplying the numerators and denominators. This gives us:

\frac{(x+3)(x+1)}{(x^2-2x-3)(x^2+2x-3)}

Factoring the Numerator and Denominator

To simplify the expression even further, we can factor the numerator and denominator. The numerator can be factored as $(x+3)(x+1)$ , and the denominator can be factored as $(x^2-2x-3)(x^2+2x-3)$ . Factoring the denominator further, we get:

(x^2-2x-3)(x^2+2x-3) = (x-3)(x+1)(x+3)(x-1)

Canceling Common Factors

Now that we have the expression in a factored form, we can cancel common factors between the numerator and denominator. The numerator has a factor of $(x+3)$ , and the denominator has a factor of $(x+3)(x-1)$ . Canceling these common factors, we get:

\frac{(x+3)(x+1)}{(x-3)(x+1)(x+3)(x-1)} = \frac{x+1}{(x-3)(x+1)(x-1)}

Canceling the Common Factor of (x+1)

We can cancel the common factor of $(x+1)$ between the numerator and denominator, leaving us with:

\frac{x+1}{(x-3)(x-1)}

Simplifying the Expression Further

We can simplify the expression further by multiplying the numerator and denominator by the reciprocal of the denominator. This gives us:

\frac{x+1}{(x-3)(x-1)} = \frac{x+1}{(x-3)(x-1)} \cdot \frac{(x-3)(x-1)}{(x-3)(x-1)} = \frac{(x+1)(x-3)(x-1)}{(x-3)^2(x-1)^2}

Canceling Common Factors Again

We can cancel common factors between the numerator and denominator, leaving us with:

\frac{(x+1)(x-3)(x-1)}{(x-3)^2(x-1)^2} = \frac{x+1}{(x-3)(x-1)}

Conclusion

In conclusion, the equivalent expression to $\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1}$ if no denominator equals zero is $\frac{x+1}{(x-3)(x-1)}$ . This expression can be obtained by using the quotient rule for division, factoring the numerator and denominator, canceling common factors, and simplifying the resulting expression.

Final Answer

The final answer is $\boxed{\frac{x+1}{(x-3)(x-1)}}$ .

Q: What is the first step in simplifying the given expression?

A: The first step in simplifying the given expression is to use the quotient rule for division, which states that division of two fractions is equivalent to multiplication by the reciprocal of the divisor.

Q: How do we apply the quotient rule for division to the given expression?

A: To apply the quotient rule for division, we multiply the first fraction by the reciprocal of the second fraction. In this case, we multiply $\frac{x+3}{x^2-2x-3}$ by $\frac{x+1}{x^2+2x-3}$ .

Q: What is the resulting expression after applying the quotient rule for division?

A: The resulting expression after applying the quotient rule for division is $\frac{(x+3)(x+1)}{(x^2-2x-3)(x^2+2x-3)}$ .

Q: How do we simplify the expression further?

A: We can simplify the expression further by factoring the numerator and denominator. The numerator can be factored as $(x+3)(x+1)$ , and the denominator can be factored as $(x^2-2x-3)(x^2+2x-3)$ .

Q: What is the factored form of the denominator?

A: The factored form of the denominator is $(x-3)(x+1)(x+3)(x-1)$ .

Q: How do we cancel common factors between the numerator and denominator?

A: We can cancel common factors between the numerator and denominator by dividing both the numerator and denominator by the common factor.

Q: What is the resulting expression after canceling common factors?

A: The resulting expression after canceling common factors is $\frac{x+1}{(x-3)(x-1)}$ .

Q: Can we simplify the expression further?

A: Yes, we can simplify the expression further by multiplying the numerator and denominator by the reciprocal of the denominator.

Q: What is the resulting expression after multiplying the numerator and denominator by the reciprocal of the denominator?

A: The resulting expression after multiplying the numerator and denominator by the reciprocal of the denominator is $\frac{(x+1)(x-3)(x-1)}{(x-3)^2(x-1)^2}$ .

Q: How do we cancel common factors between the numerator and denominator again?

A: We can cancel common factors between the numerator and denominator again by dividing both the numerator and denominator by the common factor.

Q: What is the resulting expression after canceling common factors again?

A: The resulting expression after canceling common factors again is $\frac{x+1}{(x-3)(x-1)}$ .

Q: What is the final answer?

A: The final answer is $\boxed{\frac{x+1}{(x-3)(x-1)}}$ .

Q: What is the equivalent expression to $\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1}$ if no denominator equals zero?

A: The equivalent expression to $\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1}$ if no denominator equals zero is $\frac{x+1}{(x-3)(x-1)}$ .

Q: What is the main concept used to simplify the given expression?

A: The main concept used to simplify the given expression is the quotient rule for division, factoring, and canceling common factors.

Q: What is the importance of simplifying complex fractions?

A: Simplifying complex fractions is important because it makes them easier to work with and understand. It also helps to avoid errors and makes it easier to solve problems.

Q: Can you provide an example of a real-world application of simplifying complex fractions?

A: Yes, simplifying complex fractions is used in many real-world applications, such as finance, engineering, and science. For example, in finance, simplifying complex fractions can help to calculate interest rates and investment returns. In engineering, simplifying complex fractions can help to design and optimize systems. In science, simplifying complex fractions can help to model and analyze complex systems.