Problem 75 - Prealgebra and Introductory Algebra

Question

Divide. $$\frac{y^{2}-y-56}{y^{2}+8 y+7} \div \frac{y^{2}-13 y+40}{y^{2}-4 y-5}$$

Step-by-Step Solution

Answer

\[ \frac{y+5}{y-5} \]

Step 1: Factoring the Quadratics

First, each quadratic in the numerators and denominators needs to be factored: \[ \frac{(y-8)(y+7)}{(y+7)(y+1)} \div \frac{(y-8)(y-5)}{(y+1)(y+5)} \]

Step 2: Convert Division to Multiplication

Then, rewrite the division as multiplication by taking the reciprocal of the second fraction: \[ \frac{(y-8)(y+7)}{(y+7)(y+1)} \times \frac{(y+1)(y+5)}{(y-8)(y-5)} \]

Step 3: Simplification

Now common factors in the numerator and denominator can be canceled out. Cancel \(y-8\), \(y+7\), and \(y+1\) from top and bottom to simplify: \[ \frac{1}{1} \times \frac{y+5}{y-5} \]

Step 4: Final Multiplication

Finally, because the first fraction has simplified to 1, the answer is just the second fraction: \[ \frac{y+5}{y-5} \]

Key Concepts

Factoring Quadratics

In this concept, we focus on breaking down quadratic equations into factors. A quadratic polynomial typically takes the form of \(ax^2 + bx + c\). Factoring is essentially finding two binomials that multiply to give you the original polynomial. For the problem at hand, you have several quadratics to factor.

• \(y^2 - y - 56\) turns into \((y-8)(y+7)\).
• \(y^2 + 8y + 7\) becomes \((y+7)(y+1)\).
• \(y^2 - 13y + 40\) is factored into \((y-8)(y-5)\).
• \(y^2 - 4y - 5\) results in \((y+1)(y+5)\).

Breaking down the quadratics into factors helps simplify the division problem. Each factor represents potential simplification points as you advance in solving the problem.

Fraction Multiplication

Once you have the quadratics factored, it's time to multiply fractions, a crucial step in solving complex rational expressions. Normally, to multiply two fractions, you multiply the numerators and then the denominators, resulting in a new fraction. But, dividing fractions requires an additional twist.Consider the fraction division from our exercise. Instead of dividing directly, we turn the division problem into a multiplication problem by using the reciprocal of the divisor. So, \(\frac{a}{b} \div \frac{c}{d}\) becomes \(\frac{a}{b} \times \frac{d}{c}\). For the exercise,

• Switch \(\frac{(y-8)(y-5)}{(y+1)(y+5)}\) with its reciprocal \(\frac{(y+1)(y+5)}{(y-8)(y-5)}\)

Multiplication creates a simpler pathway to cancel matching factors in the numerator and denominator.

Simplification

Simplification involves cancelling out common factors in the numerator and the denominator to reduce the expression to its simplest form. It's much like removing common terms in a fraction, just tailored to polynomials.After setting up the multiplication, observe which terms appear in both the numerator and denominator. For our division problem, note the factors:

• Cancel out \((y-8)\), \((y+7)\), and \((y+1)\) common to both the fractions.

This reduction simplifies the expression extensively, making it easier to handle. So, from our example, you simplify by switching the division to multiplication, and then cancelling equivalent factors, leading to a core simplified expression.

Reciprocal of a Fraction

Finally, let's explore the concept of reciprocals, a fundamental technique in fraction division. The reciprocal of a fraction \(\frac{a}{b}\) is simply \(\frac{b}{a}\). This form moves the denominator to the numerator position, and vice versa. While generally useful in fraction multiplication, it is indispensable when dividing fractions. The key step in transforming a division problem into a multiplication one is to take the reciprocal of the fraction that is being divided. For the given problem, the second fraction, \(\frac{(y-8)(y-5)}{(y+1)(y+5)}\), was flipped to \(\frac{(y+1)(y+5)}{(y-8)(y-5)}\) allowing for multiplication to proceed. With this approach, it not only solves the problem but illustrates the practical use of reciprocals in simplifying difficult polynomial divisions.