Problem 14 - College Algebra with Corequisite Support

Question

For the following exercises, multiply the rational expressions and express the product in simplest form. \(\frac{x^{2}-x-6}{2 x^{2}+x-6} \cdot \frac{2 x^{2}+7 x-15}{x^{2}-9}\)

Step-by-Step Solution

Answer

\(\frac{x+5}{x+3}\)

Step 1: Factor all polynomials

Factor each polynomial in the expressions. For the numerator of the first expression, \(x^2 - x - 6\), factor it to \((x-3)(x+2)\).For the denominator of the first expression, \(2x^2 + x - 6\), factor it to \((2x-3)(x+2)\).For the numerator of the second expression, \(2x^2 + 7x - 15\), factor it to \((2x-3)(x+5)\).For the denominator of the second expression, \(x^2 - 9\), factor it as the difference of squares: \((x-3)(x+3)\).

Step 2: Write factored form and multiply

After factoring, rewrite the entire expression:\[\frac{(x-3)(x+2)}{(2x-3)(x+2)} \, \cdot \, \frac{(2x-3)(x+5)}{(x-3)(x+3)}\]Multiply the numerators together and the denominators together:\[\frac{(x-3)(x+2)(2x-3)(x+5)}{(2x-3)(x+2)(x-3)(x+3)}\]

Step 3: Cancel common factors

Identify and cancel any common factors in both the numerator and the denominator:- \((x-3)\) exists in both.- \((x+2)\) exists in both.- \((2x-3)\) exists in both.After cancelling these factors, the expression simplifies to:\[\frac{(x+5)}{(x+3)}\]No further simplification is possible.

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential step when working with rational expressions. It involves breaking down a polynomial into a product of simpler polynomials, known as factors. This process makes it easier to simplify expressions later.
A polynomial like \(x^2 - x - 6\) can be factored by finding two numbers that multiply to the constant term (-6) and add to the linear coefficient (-1). These numbers are -3 and 2, so the factored form is \((x-3)(x+2)\).
For a more complex polynomial such as \(2x^2 + 7x - 15\), we use methods like grouping or trial and error to find factors. Here, it factors into \((2x-3)(x+5)\). Recognizing special formulas like the difference of squares is also crucial. For \(x^2 - 9\), which is a difference of squares, it factors to \((x-3)(x+3)\).
Learning to factor efficiently can simplify solving and manipulation of algebraic expressions. It is a skill that will benefit many areas of algebra.

Simplifying Fractions

Simplifying fractions involves reducing the expression to its simplest form. This means cancelling any common factors in the numerator and the denominator. Simplified forms are preferable as they are easier to work with and understand.
Consider a fraction representation like \(\frac{(x-3)(x+2)}{(2x-3)(x+2)}\). Here, \((x+2)\) is a common factor in the numerator and the denominator, allowing us to cancel it out. This reduces the fraction to \(\frac{x-3}{2x-3}\).
When simplifying rational expressions:

• Factor both the numerator and the denominator completely.
• Identify any common factors.
• Cancel these common factors out.

Simplification makes the fraction more elegant and easier to interpret, laying a foundation for further algebraic manipulation.

Algebraic Multiplication

Algebraic multiplication of rational expressions requires multiplying both the numerators and the denominators. Once the polynomials involved are factored, this process becomes straightforward.
Take the expressions \(\frac{(x-3)(x+2)}{(2x-3)(x+2)}\) and \(\frac{(2x-3)(x+5)}{(x-3)(x+3)}\). We multiply the numerators together and the denominators together:
Numerator: \((x-3)(x+2)(2x-3)(x+5)\)
Denominator: \((2x-3)(x+2)(x-3)(x+3)\)
This step greatly benefits from the prior factoring stage, allowing us to simply focus on combining these elements.
Finally, use simplification techniques to reduce the multiplied expression to its simplest form, which might involve cancelling out common factors between the numerator and the denominator. This results in the expression \(\frac{x+5}{x+3}\), a neat and simpler form achieved through accurate and thorough algebraic multiplication.