Problem 69 - Elementary Algebra

Question

Factor. Either factor out the greatest common factor, factor by grouping, use the guess and check method, or use the \(a c\) method. $$ 2 x^{2}+6 x+2 $$

Step-by-Step Solution

Answer

2(x^2 + 3x + 1)

Step 1: Identify the greatest common factor (GCF)

First, look for a term that can be factored out from all the terms. All the coefficients in the quadratic expression share the common factor of 2.

Step 2: Factor out the GCF

Factor 2 out from each term in the quadratic expression. This gives: \[ 2(x^2 + 3x + 1) \]

Step 3: Identify the quadratic expression to factor further

Now, focus on the quadratic expression inside the parentheses: \[ x^2 + 3x + 1 \]

Step 4: Set up the factor pairs using the ac method

For the quadratic expression \(x^2 + 3x + 1\): 1. Identify \(a = 1\), \(b = 3\), \(c = 1\) 2. Multiply \(a\) and \(c\): \[ 1 \times 1 = 1 \] 3. Find two numbers that can be multiplied to give the product (1) and that add up to give the middle coefficient (3). The pairs are \((1, 1)\), but the numbers needed to add up to 3 do not exist among whole numbers.

Step 5: Verify factor pairs or alternative factorization

Since factor pairs do not directly satisfy the requirements, focus on the factored-out expression: \[ 2(x + 3) \]

Key Concepts

Greatest Common Factor

When solving any polynomial expression, the first step often involves identifying the Greatest Common Factor (GCF). The GCF is the largest number or highest degree of a variable that divides each term of the expression without leaving a remainder. For instance, in the expression \(2x^2 + 6x + 2\), we look at each coefficient (the numbers in front of the variables) and find that 2 is the largest number that can evenly divide 2, 6, and 2. Factor out this common factor, which simplifies the expression. This gives us:
\[ 2(x^2 + 3x + 1) \]
By factoring out the GCF, you make the quadratic expression inside the parentheses simpler and easier to work with.

AC Method

The AC Method is an effective technique for factoring quadratic expressions, especially when they are not easily factorable by simpler means. This method helps to transform a quadratic expression of the form \(ax^2 + bx + c\) into a form that can be factored by grouping.

Let's break it down step by step:

• Identify the coefficients: For \(x^2 + 3x + 1\), we have \(a = 1, b = 3,\) and \(c = 1\).
• Multiply \(a\) and \(c\): Here, \(1 \times 1 = 1\).
• Find two numbers that multiply to \(ac = 1\) and add up to \(b = 3\). The pairs are (1, 1), but no pairs of whole numbers sum to 3.

Because the quadratic is simple and the middle term coefficient makes it difficult to apply the AC Method directly, it's important to verify alternative paths.
Sometimes you'll discover that, as in this example, conventional factor pairs do not satisfy all conditions.

Quadratic Expression

Quadratic expressions are polynomial expressions of degree 2, meaning they take the general form \(ax^2 + bx + c\). The standard method to factor such an expression is to rewrite it in a product of two binomials. For example, for \(x^2 + 3x + 1\), we aim to express it as \((x + p)(x + q) \) where \(p \) and \(q\) are numbers that will satisfy the conditions of sum and product.
For the quadratic expression we are working with:
1. Inside the parentheses, we have \(x^2 + 3x + 1\).
2. Notice that factoring might not yield whole numbers that fit neatly into binomials, especially if standard pairs giving the sum and product requirements do not exist.
3. In such cases, the pre-existing factors, such as the 2 we originally factored out, play a crucial role.
4. Finally, let's recall the partially factored expression: \[ 2(x^2 + 3x + 1) \]. Since we identified the difficulty in further simplifying it considering the exact factor pairs, it remains simplified using the GCF focus.