Problem 8 - Intermediate Algebra

Question

Complete each solution. $$ \begin{aligned} \frac{x^{2}-x-6}{4 x^{2}+16 x} \div \frac{x-3}{x+4} &=\frac{x^{2}-x-6}{4 x^{2}+16 x} \cdot \\ &=\frac{(x+2)(x+4)}{(x+4)(x-3)} \\ &=\frac{x+2}{\underline{\phantom{xx}}} \end{aligned} $$

Step-by-Step Solution

Answer

\( \frac{x+2}{4x} \) is the simplified expression.

Step 1: Simplify the Complex Fraction

We begin by considering the division of fractions as multiplication by the reciprocal. Thus, \( \frac{x^{2}-x-6}{4 x^{2}+16 x} \div \frac{x-3}{x+4} \) becomes \( \frac{x^{2}-x-6}{4 x^{2}+16 x} \times \frac{x+4}{x-3} \).

Step 2: Factor the Numerator and Denominator

Next, we factor the expressions where possible. The numerator \( x^2-x-6 \) factors to \( (x+2)(x-3) \). The denominator \( 4x^2 + 16x \) can be factored to \( 4x(x+4) \).

Step 3: Substitute Back into the Expression

Substitute these factors back into the expression, resulting in: \[ \frac{(x+2)(x-3)}{4x(x+4)} \times \frac{x+4}{x-3} \].

Step 4: Cancel Common Terms

We can now cancel common factors in the numerator and the denominator. The \( (x-3) \) term cancels out, and \( (x+4) \) also cancels out in both the numerator and the denominator, leaving us with \( \frac{x+2}{4x} \).

Step 5: Write the Simplified Expression

The simplified expression is \( \frac{x+2}{4x} \). This is the simplest form after cancelling the common terms in both the numerator and denominator.

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental concept in algebra that helps simplify expressions. When we factor a polynomial, we break it down into the product of its simpler components, or factors. This process can make complex algebraic expressions much easier to work with.
In the given problem, the expression \(x^2 - x - 6\) is in the numerator. To factor it, we look for two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of the middle term). These numbers are \(2\) and \(-3\). Thus, \(x^2 - x - 6\) can be factored as \((x+2)(x-3)\).
Similarly, the expression in the denominator, \(4x^2 + 16x\), can be factored by first finding a common factor in both terms. The greatest common factor is \(4x\), which when factored out, leaves \(4x(x+4)\).
Factoring transforms expressions into a multiplication of terms, simplifying the process of canceling out terms in complex fractions, for example.

Complex Fractions

Complex fractions can be intimidating at first, but they are actually straightforward when you break them down. A complex fraction is a fraction where the numerator, the denominator, or both, are themselves fractions. However, they can often be simplified by multiplying by the reciprocal.
In this exercise, we initially have the complex fraction \(\frac{x^{2}-x-6}{4 x^{2}+16 x} \div \frac{x-3}{x+4}\). To simplify this, we convert the division into multiplication by the reciprocal: \(\frac{x^{2}-x-6}{4 x^{2}+16 x} \times \frac{x+4}{x-3}\).
Once this conversion is made, it becomes easier to see opportunities for simplification, especially after factoring the polynomials involved. This change in perspective is often the first step in solving problems that involve complex fractions.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their most basic form while maintaining their equality. After factoring and rearranging terms, the goal is to cancel out any common factors in the numerator and the denominator.
In our example, after substituting the factored expressions back, the expression is \(\frac{(x+2)(x-3)}{4x(x+4)} \times \frac{x+4}{x-3}\).
By simplifying, we cancel out the common \((x-3)\) and \((x+4)\) terms from the numerator and the denominator. This cancels these terms completely, leaving the simpler expression of \(\frac{x+2}{4x}\).
The act of cancelling terms is a powerful tool in algebra, as it drastically simplifies expressions, making them easier to evaluate or use in further computations.