Problem 51 - College Algebra with Modeling and Visualization

Question

Simplify the expression. $$ \frac{x^{2}-3 x+2}{x^{2}+5 x+6} \div \frac{x^{2}+x-2}{x^{2}+2 x-3} $$

Step-by-Step Solution

Answer

The simplified expression is \(x-2\).

Step 1: Identify the Problem Type

The problem involves division of two algebraic fractions. To simplify it, we'll need to perform division of fractions by multiplying the first fraction by the reciprocal of the second.

Step 2: Write the Reciprocal

To divide the fractions, we'll convert the division into multiplication. Thus, \( \frac{x^2 - 3x + 2}{x^2 + 5x + 6} \div \frac{x^2 + x - 2}{x^2 + 2x - 3} \) becomes \( \frac{x^2 - 3x + 2}{x^2 + 5x + 6} \times \frac{x^2 + 2x - 3}{x^2 + x - 2} \).

Step 3: Factor the Polynomials

Factor each quadratic expression in the fractions:1. \(x^2 - 3x + 2\) factors to \((x-1)(x-2)\).2. \(x^2 + 5x + 6\) factors to \((x+2)(x+3)\).3. \(x^2 + x - 2\) factors to \((x-1)(x+2)\).4. \(x^2 + 2x - 3\) factors to \((x-1)(x+3)\).

Step 4: Substitute Factored Forms

Rewrite the expression using the factored forms: \[ \frac{(x-1)(x-2)}{(x+2)(x+3)} \times \frac{(x-1)(x+3)}{(x-1)(x+2)} \].

Step 5: Simplify by Cancelling Terms

Cancel common terms in the numerator and denominator:1. Cancel \(x+2\) in the denominator of the first fraction with \(x+2\) in the second denominator.2. Cancel \(x-1\) in the numerator of both fractions.3. Cancel \(x+3\) in the numerator of the second fraction with \(x+3\) in the denominator of the first fraction.This leaves \( \frac{x-2}{x-1} \).

Step 6: Final Simplified Expression

After cancelling all common factors, the simplified expression is \(x-2\).

Key Concepts

Fraction Division

Dividing fractions might initially seem tricky, but it's made simple with a straightforward rule: multiply by the reciprocal. The reciprocal of a fraction is a modification where the numerator and denominator are swapped. For example, the reciprocal of \( \frac{a}{b} \) is \( \frac{b}{a} \).
To divide one fraction by another, change the division into multiplication by taking the reciprocal of the second fraction. Consider our original expression:

• First fraction: \( \frac{x^2 - 3x + 2}{x^2 + 5x + 6} \)
• Second fraction: \( \frac{x^2 + x - 2}{x^2 + 2x - 3} \)

Conversion transforms the problem into: \( \frac{x^2 - 3x + 2}{x^2 + 5x + 6} \times \frac{x^2 + 2x - 3}{x^2 + x - 2} \).
The key is recognizing that after converting to multiplication, you maintain the same expression using factored forms.

Polynomial Factoring

Factoring polynomials is a core skill in simplifying complex algebraic expressions. It breaks down a polynomial into the product of simpler terms or factors, making them easier to work with.
For example, consider the quadratic polynomial \(x^2 - 3x + 2\). To factor it, look for two numbers that multiply to the constant term (+2) and add to the coefficient of the linear term (-3). This gives the factored form \((x-1)(x-2)\).
Other expressions in our exercise were factorized as follows:

• \(x^2 + 5x + 6\) into \((x+2)(x+3)\)
• \(x^2 + x - 2\) into \((x-1)(x+2)\)
• \(x^2 + 2x - 3\) into \((x-1)(x+3)\)

Remember, factoring helps in simplifying expressions and solving equations, as it reduces polynomials to their most basic building blocks.

Simplification of Expressions

Simplifying expressions is crucial for solving mathematical problems efficiently. After factoring the polynomials, the next step in simplification is to cancel out common factors in the numerator and the denominator.
Take the expression: \[ \frac{(x-1)(x-2)}{(x+2)(x+3)} \times \frac{(x-1)(x+3)}{(x-1)(x+2)} \].By identifying and cancelling the common factors across the numerators and denominators, you simplify the expression to:

• Cancel \((x+2)\) from both the numerator and denominator.
• Next, cancel \((x-1)\) from both parts.
• Finally, \((x+3)\) is also eliminated these common terms.

This leaves the expression as \( \frac{x-2}{x-1} \).
Such simplification not only eases calculations but also helps in gaining a deeper understanding of the mathematical structures involved.