Problem 131 - College Algebra with Modeling and Visualization

Question

Factor the expression completely. \(18 x^{2}+12 x+2\)

Step-by-Step Solution

Answer

The expression factors to \(2(3x+1)^2\).

Step 1: Identify Common Factors

First, look for any common factor in all the terms of the polynomial. The terms in the expression are 18, 12, and 2. The greatest common factor (GCF) is 2.

Step 2: Factor Out the GCF

Divide each term of the polynomial by the GCF (2) to factor it out. The expression becomes \[2(9x^2 + 6x + 1)\]

Step 3: Factor the Trinomial

Focus on factoring the trinomial inside the parentheses: \(9x^2 + 6x + 1\). Look for two numbers that multiply to \(9 \times 1 = 9\) and add to 6. These numbers are 3 and 3.

Step 4: Apply the AC Method

Rewrite the middle term (6x) using the numbers identified in the previous step: \[9x^2 + 3x + 3x + 1\]Group the terms:\[(9x^2 + 3x) + (3x + 1)\]

Step 5: Factor by Grouping

Factor out the GCD from each group. The first group, \(9x^2 + 3x\), leads to \(3x(3x + 1)\); the second group, \(3x + 1\), remains unchanged as there is no further common factor. The expression is now \[2[(3x + 1)(3x + 1)]\]

Step 6: Express in Factored Form

The expression inside the parenthesis, \((3x + 1)(3x + 1)\), can be written as \((3x + 1)^2\). Therefore, the completely factored form of the expression is \[2(3x + 1)^2\]

Key Concepts

Greatest Common Factor

The Greatest Common Factor (GCF) is a foundational concept in polynomial factoring. It represents the largest number that divides all terms of a polynomial evenly. The GCF helps simplify expressions and is the first step in various factoring techniques.

To find the GCF, list the factors of each number in the expression. For instance, in the expression \(18x^2 + 12x + 2\), consider only the numerical coefficients: 18, 12, and 2. The factors of these are:

• 18: 1, 2, 3, 6, 9, 18
• 12: 1, 2, 3, 4, 6, 12
• 2: 1, 2

The common factor in all terms is 2, which is the GCF. After identifying it, divide each term by the GCF to factor it out. This process cleans up the polynomial and prepares it for further factoring methods.

AC Method

The AC Method is a handy technique for factoring trinomials. It gets its name from the strategy of multiplying the coefficient of the first term (\(a\)) with the constant term (\(c\)). This method turns the trinomial into a simpler factorable expression.

Let's break it down using our example, \(9x^2 + 6x + 1\). First, multiply \(a\), which is 9, by \(c\), which is 1. The product is 9.

Now, the goal is to find two numbers that multiply to this product (9) and add up to \(b\), which is 6. In this case, 3 and 3 fit the description.

Rewrite the middle term (6x) using these numbers: \(9x^2 + 3x + 3x + 1\). This step transforms the trinomial into a grouping-friendly form, setting the stage for the next step in the factoring journey.

Factoring Trinomials

Factoring trinomials involves breaking down a polynomial into simpler binomial expressions, facilitating easier calculations or further mathematical operations. Once the GCF is factored out and you've applied the AC method, the expression can be rewoven into groups.

For \(9x^2 + 3x + 3x + 1\), group the terms to prepare them for factoring by grouping: \((9x^2 + 3x) + (3x + 1)\). Observe each grouping separately.

In the first group, \(9x^2 + 3x\), factor out the common element \(3x\), resulting in \(3x(3x + 1)\). The second group, \(3x + 1\), surprisingly holds no common factor, leaving it unchanged.

The expression is now organized as \(2[(3x + 1)(3x + 1)]\). Through factoring trinomials, the polynomial is decomposed into \(2(3x + 1)^2\), a neat and completely factored form.