Problem 85 - Intermediate Algebra

Question

Simplify each expression. If an expression cannot be simplified, write "Does not simplify." $$ \frac{2 x^{2}+2 x-12}{x^{3}+3 x^{2}-4 x-12} $$

Step-by-Step Solution

Answer

\(\frac{2}{x+2}\), where \(x \neq -3\) and \(x \neq 2\).

Step 1: Factor the Numerator

First, let's factor the numerator, which is \(2x^2 + 2x - 12\). Start by factoring out the greatest common factor (GCF): \(2(x^2 + x - 6)\). Now, we need to factor the quadratic \(x^2 + x - 6\). We'll find two numbers that multiply to \(-6\) and add to \(1\), which are \(3\) and \(-2\). Therefore, the numerator factors as \(2(x + 3)(x - 2)\).

Step 2: Factor the Denominator

Next, we'll factor the denominator \(x^3 + 3x^2 - 4x - 12\). Start by using the factor by grouping method. Group the terms: \((x^3 + 3x^2) + (-4x - 12)\). Factor each group: \(x^2(x + 3) - 4(x + 3)\). Now, factor out the common factor \((x + 3)\) to get \((x + 3)(x^2 - 4)\). The expression \(x^2 - 4\) is a difference of squares, so it can be further factored as \((x + 2)(x - 2)\). Thus, the denominator becomes \((x + 3)(x + 2)(x - 2)\).

Step 3: Simplify the Expression

The expression is now \(\frac{2(x + 3)(x - 2)}{(x + 3)(x + 2)(x - 2)}\). There are common factors in the numerator and the denominator. Specifically, \((x + 3)\) and \((x - 2)\) can be cancelled out from both the numerator and the denominator. This simplifies the expression to \(\frac{2}{x + 2}\), where \(x eq -3\) and \(x eq 2\) to prevent division by zero in the original expression.

Key Concepts

Factoring Polynomials

Polynomials are algebraic expressions that consist of variables and coefficients. Simplifying them often requires factoring, which means breaking down the polynomial into products of simpler polynomials. Consider the numerator in the expression: \[ 2x^2 + 2x - 12 \]. To factor this, start by identifying the greatest common factor (GCF), which is 2. Factor out the GCF to get:

• \( 2(x^2 + x - 6) \)

Then, look at the quadratic expression \( x^2 + x - 6 \). Find two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(3\) and \(-2\), which can further factor the quadratic into:

• \( (x + 3)(x - 2) \)

The polynomial initially was overly complex, but through factoring, it transforms into \[ 2(x + 3)(x - 2) \], which is more manageable.

Greatest Common Factor

The Greatest Common Factor (GCF) in an expression is the largest factor that divides all terms of it. Finding the GCF is usually the first step in factoring since it simplifies the polynomial, making further factoring less complex. Take the term \( 2x^2 + 2x - 12 \) from the numerator. Each term of the polynomial— \(2x^2\), \(2x\), and \(-12\) —shares a common factor of 2. Factoring out the GCF first simplifies the expression into:

• \( 2(x^2 + x - 6) \)

This action reduces the complexity and makes it easier to handle the inside terms \( x^2 + x - 6 \). Once refactored, further simplification follows. GCF helps in reducing problems into simpler, more approachable segments, which is crucial for effective mathematical problem-solving.

Difference of Squares

The difference of squares is a special factoring pattern, used when a quadratic is expressed as: \[ a^2 - b^2 \]. This pattern factors into products as \( (a + b)(a - b) \).When analyzing the denominator of the given expression, we discover \( x^2 - 4 \). Recognize this as \( x^2 - 2^2 \), which clearly fits the difference of squares format. Applying the difference of squares pattern transforms \( x^2 - 4 \) into:

• \( (x + 2)(x - 2) \)

This approach is powerful and simplifies expressions where square terms subtract from each other. Integers, and more specifically square integers, help identify and utilize this pattern, which speeds up the simplification of algebraic expressions systematically.