Problem 65 - A Graphical Approach to College Algebra

Question

Completely factor each polynomial by substitution. $$6(4 z-3)^{2}+7(4 z-3)-3$$

Step-by-Step Solution

Answer

The completely factored form is \((12z - 10)(8z - 3) \).

Step 1: Substitute for Simplicity

Let \( x = 4z - 3 \). This turns the original polynomial \( 6(4z-3)^2 + 7(4z-3) - 3 \) into \( 6x^2 + 7x - 3 \).

Step 2: Factor the Quadratic

To factor \( 6x^2 + 7x - 3 \), find two numbers that multiply to \( 6 \times (-3) = -18 \) and add to \( 7 \). The numbers are \( 9 \) and \( -2 \). Rewrite \( 7x \) as \( 9x - 2x \):\[ 6x^2 + 9x - 2x - 3 \]

Step 3: Group and Factor

Group the terms: \( (6x^2 + 9x) + (-2x + 3) \). Factor each group:\[ 3x(2x + 3) - 1(2x + 3) \]

Step 4: Factor Out Common Terms

Factor out the common term \((2x + 3)\):\[ (3x - 1)(2x + 3) \]

Step 5: Substitute Back the Original Expression

Replace \( x \) with \( 4z - 3 \) to convert back to the variable \( z \):\[ (3(4z - 3) - 1)(2(4z - 3) + 3) \]

Step 6: Simplify Each Factor

Simplify each factor:1. \( 3(4z - 3) - 1 = 12z - 9 - 1 = 12z - 10 \)2. \( 2(4z - 3) + 3 = 8z - 6 + 3 = 8z - 3 \)The completely factored form is \( (12z - 10)(8z - 3) \).

Key Concepts

Quadratic Equations

A quadratic equation is a fundamental concept in algebra, typically represented in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations take the shape of a parabola when graphed. They can have:

• Two real roots
• One real root
• No real roots (but two complex roots)

Quadratic equations emerge frequently in various mathematical contexts, including physics and finance, where they model projectile motions or calculate interest rates. Understanding how to solve quadratic equations is crucial because they are everywhere. Methods for solving them include:

• Factoring
• Using the Quadratic Formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
• Completing the Square

Factoring is often the most intuitive method when the polynomial can be easily decomposed into two binomials. Identifying two numbers that add to the coefficient of \( x \) and multiply to the product of the leading coefficient and the constant helps in the factoring method. This technique is fundamental in algebra and invaluable in solving quadratic equations.

Algebraic Substitution

Algebraic substitution is a powerful technique used to simplify complex equations or expressions by replacing variables with other expressions. This makes them easier to work with. In our original exercise, substitution helps manage a more complicated polynomial by turning it into a simpler quadratic equation.
Here's how substitution generally works:

• Choose a new variable, often \( x \), to represent a more complex part of the expression.
• Make the substitution to simplify the equation.
• Perform operations on the simpler equation.
• Once simplified or solved, substitute back the original expression.

For example, in the exercise:- We use \( x = 4z - 3 \). - The polynomial \( 6(4z-3)^2 + 7(4z-3) - 3 \) simplifies to \( 6x^2 + 7x - 3 \).- After solving the quadratic equation, we substitute \( 4z - 3 \) back to express the final solution in terms of the original variable, \( z \).
Substitution is convenient when dealing with nested functions or when an expression repeatedly appears. It aids in reducing potential algebraic mistakes and provides clarity.

Factoring by Grouping

Factoring by grouping is a method used to factor polynomials that do not have obvious factorizations like quadratics with no constant. It involves rearranging and grouping terms in a way that makes them easier to factor.
To perform factoring by grouping, follow these steps:

• Split the middle term (if any) into two separate terms whose product equals the product of the outer coefficients, i.e., the leading coefficient and the constant term.
• Group terms into two pairs so that each pair can have a common factor.
• Factor out the common factor from each pair.
• If successful, the factored expression will reveal a common binomial factor.

In the given problem, once we've substituted \( x \) for \( 4z-3 \), the polynomial becomes \( 6x^2 + 7x - 3 \). By:1. Splitting \( 7x \) into \( 9x - 2x \), we can group:2. \( (6x^2 + 9x) + (-2x - 3) \).3. Factoring each group gives \( 3x(2x + 3) - 1(2x + 3) \).4. Factoring out the common \( (2x + 3) \) gives \( (3x - 1)(2x + 3) \).
Thus, factoring by grouping provides a structured approach to solving complex factorization problems by breaking them into manageable steps.