Problem 18 - Elementary Algebra

Question

Factor. $$ 12 x 2-88 x-15 $$

Step-by-Step Solution

Answer

The factored form is \((6x + 1)(2x - 15)\).

Step 1: Identify the expression to factor

The given expression is \( 12x^2 - 88x - 15 \). Our task is to factor this quadratic expression.

Step 2: Check for common factors

Examine the coefficients (12, -88, -15) to see if there is a greatest common factor (GCF) for all terms. There is no GCF other than 1, so we proceed without factoring any common factor.

Step 3: Use the AC method

The expression is in the form \( ax^2 + bx + c \), where \( a = 12 \), \( b = -88 \), and \( c = -15 \). Multiply \( a \) and \( c \): \( 12 \times -15 = -180 \). We need to find two numbers that multiply to -180 and add to -88.

Step 4: Find the numbers that work

The numbers -90 and 2 multiply to -180 and add to -88. These are the numbers we need to split the middle term.

Step 5: Rewrite the middle term and factor by grouping

Rewrite \( 12x^2 - 88x - 15 \) as \( 12x^2 - 90x + 2x - 15 \). Group the terms: \((12x^2 - 90x) + (2x - 15)\).

Step 6: Factor each group

Factor out the GCF from each group. For \( 12x^2 - 90x \), the GCF is 6x, giving \( 6x(2x - 15) \). For \( 2x - 15 \), the GCF is 1, giving \( 1(2x - 15) \).

Step 7: Complete the factorization

The expression is now factored as \( (6x + 1)(2x - 15) \).

Key Concepts

Understanding the Greatest Common Factor (GCF)

When we factor a quadratic expression, the first step is to check if all the terms share a common factor, known as the Greatest Common Factor (GCF). The GCF is the largest number that divides each of the coefficients in the expression evenly. In simpler terms, it's the biggest number we can "pull out" of each term.

For example, in the expression \(12x^2 - 88x - 15\), we focus on the numerical coefficients \((12, -88, -15)\). Here:

• 12 factors into \(2 \times 2 \times 3\).
• 88 factors into \(2 \times 2 \times 2 \times 11\).
• 15 factors into \(3 \times 5\).

Checking these, we can't find any number other than 1 that divides all three coefficients equally, so we conclude that there's no GCF to factor out from \(12x^2 - 88x - 15\).

Even though no GCF other than 1 exists, considering this step ensures no potential factor is overlooked, allowing for neat and precise factorization later.

Applying the AC Method

The AC method is a common technique used for factoring quadratics when a simple GCF isn't available. For a quadratic in the form \(ax^2 + bx + c\), the process begins by multiplying \(a\) and \(c\).

In our expression \(12x^2 - 88x - 15\), the coefficients are \(a = 12\), \(b = -88\), and \(c = -15\). Here’s how we apply the AC method:

• Calculate \(a \times c\), which gives us \(12 \times -15 = -180\).
• Find two numbers that multiply to \(-180\) and add to the middle term \((-88)\).

These two numbers form the basis of rewriting the middle term, ensuring it captures both multiplicative and additive properties required for successful factorization. In this case, the numbers \((-90)\) and \(2\) work perfectly, as they multiply to \(-180\) and sum to \(-88\).

Using the AC method streamlines the factoring process, helping break down complex expressions into manageable parts for easier factor grouping.

Factor by Grouping

Once you know the numbers that work from the AC method, the next step is to rewrite the expression and perform factor by grouping. This tactic involves reorganizing and grouping terms in the expression to facilitate the process of finding common factors.

For our example, we split the middle term \(-88x\) into \(-90x + 2x\), giving us \(12x^2 - 90x + 2x - 15\). Here's the step-by-step of factor by grouping:

• Group terms: \((12x^2 - 90x) + (2x - 15)\).
• Factor out the greatest common factor from each group. From \(12x^2 - 90x\), we factor out \(6x\), resulting in \(6x(2x - 15)\). From \(2x - 15\), we factor out \(1\), to get \(1(2x - 15)\).

Notice that \((2x - 15)\) appears in both groups, thus becoming a common factor. This allows us to combine both parts into \((6x+1)(2x-15)\), achieving full factorization.

Factor by grouping turns potentially complicated factorization into a series of simpler, intuitive steps, emphasizing the importance of organizing the expression to reveal its factors clearly.