Problem 24 - College Algebra

Question

Divide the rational expressions. $$ \frac{3 y^{2}-7 y-6}{2 y^{2}-3 y-9} \div \frac{y^{2}+y-2}{2 y^{2}+y-3} $$

Step-by-Step Solution

Answer

\( \frac{(3y + 2)(2y - 3)(y + 1)}{(2y + 3)(y + 2)(y - 1)} \) after simplification.

Step 1: Understand Division of Rational Expressions

To divide one fraction by another, we multiply by the reciprocal of the second fraction. That is, we convert \( a \div b \) into \( a \cdot \frac{1}{b} \).

Step 2: Write the Division as Multiplication of Reciprocals

Convert the given problem into a multiplication problem: \[ \frac{3 y^{2}-7 y-6}{2 y^{2}-3 y-9} \cdot \frac{2 y^{2}+y-3}{y^{2}+y-2} \].

Step 3: Factor Each Quadratic Expression

Factor each of the quadratic expressions:- \(3y^2 - 7y - 6\) factors to \((3y + 2)(y - 3)\).- \(2y^2 - 3y - 9\) factors to \((2y + 3)(y - 3)\).- \(y^2 + y - 2\) factors to \((y + 2)(y - 1)\).- \(2y^2 + y - 3\) factors to \((2y - 3)(y + 1)\).

Step 4: Substitute the Factors Back

Replace each quadratic expression in the multiplication problem with its factored form:\[ \frac{(3y + 2)(y - 3)}{(2y + 3)(y - 3)} \cdot \frac{(2y - 3)(y + 1)}{(y + 2)(y - 1)} \].

Step 5: Cancel Common Factors

Identify and cancel out common factors between numerators and denominators. The \( (y - 3) \) term is a common factor:\[ \frac{(3y + 2)}{(2y + 3)} \cdot \frac{(2y - 3)(y + 1)}{(y + 2)(y - 1)} \].

Step 6: Multiply the Remaining Expressions

Multiply the remaining factors:\[ \frac{(3y + 2)(2y - 3)(y + 1)}{(2y + 3)(y + 2)(y - 1)} \].

Step 7: Write the Simplified Expression

The simplified expression is \( \frac{(3y + 2)(2y - 3)(y + 1)}{(2y + 3)(y + 2)(y - 1)} \).

Key Concepts

Division of Rational Expressions

When working with division of rational expressions, the key is to remember that dividing by a fraction is the same as multiplying by its reciprocal. This means turning a division problem into a multiplication problem can simplify things greatly. Let's break it down:

• Rational expressions are like fractions that have polynomials as numerators and denominators.
• To divide one rational expression by another, multiply the first by the reciprocal (or flipped version) of the second.

For example, in our exercise, we've rewritten \[ \frac{3 y^{2}-7 y-6}{2 y^{2}-3 y-9} \div \frac{y^{2}+y-2}{2 y^{2}+y-3} \]as:\[ \frac{3 y^{2}-7 y-6}{2 y^{2}-3 y-9} \cdot \frac{2 y^{2}+y-3}{y^{2}+y-2} \].By doing this, we set the stage for simpler operations, like factoring and cancelling, that lead us to the solution.

Factoring Quadratic Expressions

Factoring plays a crucial role in simplifying rational expressions. It involves breaking down complex expressions into products of simpler expressions. Quadratic expressions follow the pattern of \( ax^2 + bx + c \), and factoring them can sometimes be tricky. Here's a simple method:

• Find two numbers that multiply to \( a \times c \) and add to \( b \).
• Use these numbers to rewrite the middle term and factor by grouping if needed.

In the exercise, we factored each quadratic expression like so:

• \(3y^2 - 7y - 6\) becomes \((3y + 2)(y - 3)\).
• \(2y^2 - 3y - 9\) becomes \((2y + 3)(y - 3)\).
• \(y^2 + y - 2\) becomes \((y + 2)(y - 1)\).
• \(2y^2 + y - 3\) becomes \((2y - 3)(y + 1)\).

By converting these complex expressions into factors, we make the upcoming operations of multiplying and cancelling much easier.

Multiplication of Rational Expressions

Once the division problem is rewritten as a multiplication task, the process becomes much like multiplying two fractions. Here's how to handle multiplication of rational expressions:

• Multiply across the numerators to find the new numerator.
• Multiply across the denominators to find the new denominator.

Using the factored forms of the quadratic expressions we found earlier, we multiply them like so:\[ \frac{(3y + 2)(y - 3)}{(2y + 3)(y - 3)} \cdot \frac{(2y - 3)(y + 1)}{(y + 2)(y - 1)} \]This results in a long expression with all factors multiplied together:\[ \frac{(3y + 2) \cdot (2y - 3) \cdot (y + 1)}{(2y + 3) \cdot (y + 2) \cdot (y - 1)} \].This step sets us up for the final simplification by cancelling common factors.

Cancelling Common Factors

Cancelling common factors is the ultimate step that leads to the simplification of the expression. It aims to cross out the same terms found in the numerator and the denominator, reducing fractions to their simplest form.

• Identify any terms that repeat in both the numerator and the denominator.
• Cancel these terms by dividing them out since they equal 1 when divided by themselves.

In this exercise, we noticed that \((y - 3)\) is a common factor in the numerator and denominator for one expression:\[ \frac{(3y + 2)}{(2y + 3)} \cdot \frac{(2y - 3)(y + 1)}{(y + 2)(y - 1)} \].Now, multiply what's left, and you'll have your simplified expression:\[ \frac{(3y + 2)(2y - 3)(y + 1)}{(2y + 3)(y + 2)(y - 1)} \].Through cancelling, the expression is much simpler and easier to interpret, providing a more elegant and concise solution.