Problem 31 - Algebra A Combined Function

Question

Factor each trinomial by grouping. Exercises 9 through 12 are broken into parts to help you get started. $$ -13 x+6+6 x^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Write in Standard Form

Reorder the given expression so the terms are in descending order of the exponents. The expression becomes \(6x^2 - 13x + 6\).

Step 2: Multiply and Identify

Identify the product of the leading coefficient (6) and the constant term (6). So, \(6 \cdot 6 = 36\). We need two numbers that multiply to 36 and add to the middle coefficient, -13.

Step 3: Find Suitable Pair

Find two numbers that multiply to 36 and add up to -13. These numbers are -9 and -4, since \((-9) \times (-4) = 36\) and \((-9) + (-4) = -13\).

Step 4: Rewrite the Middle Term

Rewrite the middle term, -13x, using the numbers found: \(6x^2 - 9x - 4x + 6\).

Step 5: Group Terms

Group the terms in pairs: \((6x^2 - 9x) + (-4x + 6)\).

Step 6: Factor Each Group

Factor each group separately. In the first group, factor out 3x: \(3x(2x - 3)\). In the second group, factor out -2: \(-2(2x - 3)\).

Step 7: Factor by Grouping

Both expressions have a common factor of \((2x - 3)\). Thus, the expression becomes \((3x - 2)(2x - 3)\).

Step 8: Final Verification

Verify the factored form by expanding \((3x - 2)(2x - 3)\) to ensure it equals the original expression, \(6x^2 - 13x + 6\).

Key Concepts

Polynomial Expressions

Polynomial expressions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. For example, the expression \(6x^2 - 13x + 6\) is a polynomial with three terms, commonly referred to as a trinomial.
Trinomials, specifically, are polynomial expressions with exactly three terms, like our given example. Understanding polynomial expressions involves recognizing their structure.

• The number in front of a term is the coefficient, such as 6 in \(6x^2\).
• The variable is the letter, in this case, \(x\).
• The exponent is the power to which the variable is raised, with 2 being the exponent in the term \(6x^2\).
• The expression is called a quadratic if its highest exponent is 2.

These components help us determine how to manipulate the expression, such as factoring by grouping, to simplify or solve for specific values.

Factoring by Grouping

Factoring by grouping is a powerful technique used to simplify polynomial expressions, particularly trinomials, by breaking them down into simpler terms that can be factored further. The goal is to identify pairs of terms within the polynomial expression that share common factors.
In our example, we begin with the trinomial \(6x^2 -13x + 6\). The process starts by rewriting the middle term in a way that introduces numbers that add up to the middle coefficient (-13) and multiply to the product of the leading coefficient (6) and the constant term (6), i.e., 36.
In this case, the numbers -9 and -4 achieve the needed condition:

• Multiply: \((-9) \times (-4) = 36\)
• Add up: \((-9) + (-4) = -13\)

With these numbers, rewrite the trinomial as \(6x^2 - 9x - 4x + 6\). Then group the terms: \((6x^2 - 9x) + (-4x + 6)\).
Factor out common factors from each group:

• \(3x(2x - 3)\) from the first group.
• \(-2(2x - 3)\) from the second group.

Finally, recognize the common binomial \((2x - 3)\) and write the complete factored expression as \((3x - 2)(2x - 3)\). This structured approach, known as factoring by grouping, simplifies the polynomial into its root factors.

Quadratic Equations

Quadratic equations are equations that can be expressed in the standard polynomial form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(x\) is the variable. These equations are pivotal in algebra because they can model a wide range of real-world phenomena.
The trinomial \(6x^2 - 13x + 6\), once factored into \((3x - 2)(2x - 3)\), can be set equal to zero to find solutions for \(x\). This is the main utility of factoring quadratic equations.
To solve, set each factor to zero:

• \(3x - 2 = 0\) which gives \(x = \frac{2}{3}\).
• \(2x - 3 = 0\) which gives \(x = \frac{3}{2}\).

These solutions are the roots of the equation, representing the values of \(x\) that satisfy the original polynomial equation equal to zero. Factoring, as seen in this exercise, is one of the most efficient methods for solving quadratic equations when they are factorable.