Problem 51 - Elementary and Intermediate Algebra

Question

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 6 x^{2}-14 x+8 $$

Step-by-Step Solution

Answer

The factored form is \(2(3x-4)(x-1)\).

Step 1: Identify Common Factors

Start by checking if there are any common factors in all the terms of the polynomial \(6x^2 - 14x + 8\). In this case, all the coefficients do not have a common factor greater than 1.

Step 2: Factor Using the AC Method

The expression is in the form \(ax^2 + bx + c\) with \(a = 6\), \(b = -14\), \(c = 8\). Multiply \(a\) and \(c\) to get \(6 \times 8 = 48\). We need two numbers that multiply to 48 and add to -14. These numbers are -8 and -6.

Step 3: Rewrite the Middle Term

Rewrite \(-14x\) as \(-8x - 6x\) so the expression becomes \(6x^2 - 8x - 6x + 8\).

Step 4: Group Terms

Group the terms: \((6x^2 - 8x) + (-6x + 8)\).

Step 5: Factor by Grouping

Factor out the common factors in each group. \[6x^2 - 8x = 2x(3x - 4)\] and \[-6x + 8 = -2(3x - 4)\]. The expression is now \(2x(3x - 4) - 2(3x - 4)\).

Step 6: Factoring Completely

Since \((3x - 4)\) is common in both terms, factor it out: \((3x - 4)(2x - 2)\).

Step 7: Simplify

Notice the term \((2x - 2)\) can be further simplified by factoring out a 2, resulting in \((3x - 4)(2)(x - 1)\). The completely factored form is \(2(3x - 4)(x - 1)\).

Key Concepts

AC Method

The AC method is a popular technique for factoring quadratic expressions, especially when the leading coefficient is not equal to 1. The method's name comes from the product of the coefficients 'a' and 'c' in a quadratic expression formatted as \(ax^2 + bx + c\). This method is crucial when solving polynomials that cannot be easily factored by inspection.

To use the AC method, follow these simple steps:

• Multiply the first and last coefficients, that is, \(a\cdot c\).
• Look for two numbers that multiply to \(ac\) and sum to the middle coefficient \(b\).
• Use these two numbers to split the middle term into two separate terms, then rewrite the expression.
• Proceed with factoring by grouping the newly formatted expression.

For instance, consider our original expression, \(6x^2 - 14x + 8\). We first multiply \(6\) and \(8\) to get \(48\). Our task is then to find two numbers that multiply to \(48\) and additionally combine to give \(-14\). The numbers \(-8\) and \(-6\) fit perfectly. After rewriting \(-14x\) with these two numbers, it makes the expression ready for the next step, factoring by grouping.

Factoring by Grouping

Factoring by grouping is a strategic method used when a polynomial has four or more terms. Using the AC method, the expression is transformed into one suitable for grouping. This technique plays a vital role in simplifying the polynomial through logical organization.

Here's how to efficiently group and factor the expression:

• After rewriting the expression, organize it into two distinct pairs.
• Factor out the greatest common factor within each pair.
• Recognize the common binomial factor in the two groups resulting from factoring.
• Factor out the common binomial.

For example, after rewriting \(-14x\) as \(-8x - 6x\), the expression becomes \(6x^2 - 8x - 6x + 8\). By grouping \(6x^2 - 8x\) and \(-6x + 8\), we can factor them to \(2x(3x - 4)\) and \(-2(3x - 4)\), respectively. Notice the common binomial \((3x - 4)\), which can be factored out to simplify the expression further.

Quadratic Expressions

Quadratic expressions are polynomials of the form \(ax^2 + bx + c\) where \(a, b,\) and \(c\) are constants. Understanding how to factor these expressions is foundational in algebra, as they frequently appear in equations, functions, and graphs.

Essentials about quadratic expressions:

• The term \(ax^2\) is the quadratic term, representing the highest degree and indicating the parabolic dimension of the expression.
• The linear term \(bx\) influences the symmetry axis of a corresponding parabola.
• The constant term \(c\) impacts the vertical position or the y-intercept of a graph.

Successfully factoring quadratic expressions involves breaking them down into simpler linear terms involving their roots. In our example, we see \(6x^2 - 14x + 8\) factored into \(2(3x - 4)(x - 1)\). This not only simplifies problem solving but also aids in graphing and finding solutions to quadratic equations. By mastering these, students can confidently tackle more complex algebraic problems.