Problem 55 - Elementary and Intermediate Algebra

Question

Perform the operations and simplify, if possible. See Example 5 $$\left(x^{2}+x-2 c x-2 c\right) \cdot \frac{x^{2}+3 x+2}{4 c^{2}-x^{2}}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{(x-2c)(x+2)}{(2c+x)(2c-x)}\).

Step 1: Simplify each polynomial expression

Begin by simplifying any expressions that can be factored. - For \(x^2 + x - 2cx - 2c\), group and factor to get: \((x^2 + x) - (2cx + 2c)\), which further simplifies to \(x(x+1) - 2c(x+1) = (x-2c)(x+1)\).- For \(x^2 + 3x + 2\), factor as \((x + 1)(x + 2)\). - For \(4c^2 - x^2\), recognize it as a difference of squares: \((2c + x)(2c - x)\).

Step 2: Rewrite the expression with factored forms

Substitute the factored expressions into the original product: \[((x-2c)(x+1)) \cdot \frac{(x+1)(x+2)}{(2c+x)(2c-x)}\]

Step 3: Cancel common factors in the expression

Identify and cancel factors that appear in both the numerator and denominator. - The factor \((x+1)\) is in both the numerator and denominator. Remove \((x+1)\) from both: \[(x-2c) \cdot \frac{x+2}{(2c+x)(2c-x)}\]

Step 4: Write the simplified expression

After canceling, the expression simplifies to: \[\frac{(x-2c)(x+2)}{(2c+x)(2c-x)}\] Now, all polynomials are fully factored and the expression is simplified.

Key Concepts

Factoring Polynomials

Factoring polynomials involves breaking down the polynomial into a product of simpler polynomials. This is a key step in polynomial operations, as it simplifies the expression and makes it easier to work with.
For instance, consider the polynomial \(x^2 + x - 2cx - 2c\). The first step in factoring is to group terms that can be factored together. Here, we can group terms as \((x^2 + x) - (2cx + 2c)\). Notice the common term in each group: \((x + 1)\) and \(2c\). Factor each group to get \((x(x+1) - 2c(x+1))\), which further simplifies to \((x - 2c)(x + 1)\).
Another polynomial, \(x^2 + 3x + 2\), factors into \((x + 1)(x + 2)\). This is a straightforward case because you can find two numbers that multiply to 2 and add to 3, namely 1 and 2. This process sometimes requires trial and error but becomes more intuitive with practice.
Effective factoring reduces complicated problems into simpler ones that are easier to solve.

Difference of Squares

The difference of squares is a common polynomial pattern that simplifies expressions significantly. It is recognizable by its form: \(a^2 - b^2\), and can be factored into \((a - b)(a + b)\). This pattern occurs because the terms cancel each other out when expanded. For example:

• \(a^2 - b^2 = (a - b)(a + b)\)
• When expanded: \((a - b)(a + b) = a^2 - b^2\)

In the expression \(4c^2 - x^2\), this pattern is clear. Here, \(4c^2\) is a perfect square because it can be written as \((2c)^2\), and \(x^2\) is also a perfect square. Therefore, \(4c^2 - x^2\) can be factored as \((2c + x)(2c - x)\).
Recognizing this structure allows us to streamline complex polynomials into manageable factors, which can then be simplified further when combined with other expressions.

Simplifying Rational Expressions

Simplifying rational expressions akin to simplifying fractions: you reduce to the simplest form by canceling common factors in the numerator and denominator.
In the provided exercise, after factoring each polynomial, we rewrite the expression as:

• \[((x-2c)(x+1)) \/ (2c+x)(2c-x)\]
• \((x-2c) \/ \frac{x+2}{(2c+x)(2c-x)}\)

The key is identifying and canceling common factors. Here, the common factor \((x+1)\) shows up in both the numerator \((x-2c)(x+1)(x+2)\) and in the denominator \((2c+x)(2c-x)\). By canceling \((x+1)\) from both sides, the expression condenses to:

• \[\frac{(x-2c)(x+2)}{(2c+x)(2c-x)}\]

Simplifying rational expressions involves not only mastering polynomial factoring but also recognizing shared elements across different parts of the expression. This skill reduces complexity, shows equivalent expressions, and often highlights potential solutions or causes for undefined behavior in equations.