Problem 52 - College Algebra with Modeling and Visualization

Question

Simplify the expression. $$ \frac{2 x^{2}+x-1}{6 x^{2}+x-2} \div \frac{2 x^{2}+5 x+3}{6 x^{2}+13 x+6} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Division of Fractions

The given problem involves dividing two fractions. Generally, to divide by a fraction, you multiply by its reciprocal. This means we can express this division as a multiplication problem: \( \frac{2x^2 + x - 1}{6x^2 + x - 2} \times \frac{6x^2 + 13x + 6}{2x^2 + 5x + 3} \).

Step 2: Factor the Polynomials in the Numerator and Denominator

To simplify the multiplication of these rational expressions, factor each of the quadratic expressions if possible.1. \(2x^2 + x - 1 = (2x - 1)(x + 1)\)2. \(6x^2 + x - 2 = (3x - 2)(2x + 1)\)3. \(2x^2 + 5x + 3 = (2x + 3)(x + 1)\)4. \(6x^2 + 13x + 6 = (3x + 2)(2x + 1)\)Ensure the factors are correct by expanding them back to check.

Step 3: Rewrite the Expression Using Factored Form

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

Step 4: Simplify the Expression by Cancelling Common Factors

Look for common factors in the numerator and denominator across the entire expression. Here, \((2x + 1)\) cancels with \((2x + 1)\) and \((x + 1)\) cancels with \((x + 1)\). The expression simplifies to: \[\frac{(2x - 1)(3x + 2)}{(3x - 2)(2x + 3)}\].

Step 5: Present the Simplified Expression

The expression is now fully simplified. Ensure that no more factors can be cancelled. The final expression is:\[\frac{2x - 1}{3x - 2} \times \frac{3x + 2}{2x + 3} = \frac{(2x - 1)(3x + 2)}{(3x - 2)(2x + 3)}\].

Key Concepts

Factoring Polynomials

Factoring polynomials plays a crucial role when working with rational expressions. To factor a polynomial means to express it as a product of simpler polynomials. This process is key to simplifying expressions because it allows you to identify and cancel common factors.

Consider a quadratic polynomial, typically given in the form \(ax^2 + bx + c\). The goal is to find two binomials \((px + q)(rx + s)\) that multiply to give the original polynomial. This involves:

• Finding two numbers that multiply to \(ac\) (the product of the leading coefficient and the constant term)
• Ensuring these numbers add up to \(b\) (the coefficient of the middle term)

Once these numbers are determined, the original polynomial can be split and rearranged, making factoring possible. In our exercise, each polynomial was factored into its binomial components, which later allows for simplification.

Simplifying Expressions

Simplifying expressions is about making them as straightforward as possible without changing their value. For rational expressions, this means canceling any common factors between the numerator and the denominator.

Once a rational expression is factored completely, simplification can take place by canceling these common factors crosswise between the numerator and the denominator. It’s important to factor both the numerators and the denominators completely to identify any common components. In our exercise, after factoring, the expressions were rewritten in their simplest form by canceling terms like \((2x + 1)\) and \((x + 1)\), which appeared in both the numerator and denominator.

Division of Fractions

The division of fractions can initially seem daunting, but it follows a straightforward rule: multiply by the reciprocal. When dividing fractions, you essentially flip (or "invert") the second fraction and change the division sign to multiplication.

For the problem at hand, this is shown by changing the expression \(\frac{2x^2 + x - 1}{6x^2 + x - 2} \div \frac{2x^2 + 5x + 3}{6x^2 + 13x + 6}\) into a multiplication problem: \(\frac{2x^2 + x - 1}{6x^2 + x - 2} \times \frac{6x^2 + 13x + 6}{2x^2 + 5x + 3}\).

Once this conversion is made, the next steps often involve factoring and simplifying as discussed in previous sections. Understanding this basic rule is vital, as it forms the foundation for tackling more complex rational expressions.