Problem 13 - College Algebra

Question

Solve each equation in Exercises \(1-14\) by factoring. $$7-7 x=(3 x+2)(x-1)$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = -3\) and \(x = 1\).

Step 1: Simplify and Rearrange the Equation

Firstly, distribute the values on the right side of the equation: \(3x * x = 3x^2\) and \(3x * -1 = -3x\) while \(2 * x = 2x\) and \(2 * -1 = -2\). After distributing, the equation becomes: \(7-7x = 3x^2 - x - 2\). Rearrange the equation to bring everything to one side. By doing this we have: \(3x^2 - x - 2 - 7 +7x = 0\) or simplified \(3x^2 + 6x - 9 = 0\).

Step 2: Factorize the Quadratic Equation

Now, factorize the equation \(3x^2 + 6x - 9 = 0\). By looking for two numbers that multiply to -27 (i.e. 3 * -9) and add up to 6. The numbers that fit this criterion are 9 and -3. We can rewrite the equation as: \(3x^2 + 9x - 3x - 9 = 0\). Now, group these terms to find common factors: \(3x(x + 3) - 3(x + 3)=0\). We now factor out the common factor \(x + 3\) to get \((x + 3)(3x - 3) = 0\).

Step 3: Solve for the variable 'x'

Lastly, solve for 'x' by setting each factor equal to zero. Thus, \(x + 3 = 0\) and \(3x - 3 = 0\). Solving the first equation gives \(x = -3\) and the second equation gives \(x = 1\). Therefore, the solutions of the equation are \(x = -3\) and \(x = 1\).

Key Concepts

Quadratic Equation

Understanding quadratic equations is essential in algebra. A quadratic equation is a type of polynomial equation of degree 2. The general form is expressed as \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, and \(a eq 0\). This equation graphically represents a parabola when plotted on a graph.

In the solution we analyzed, the quadratic equation was given initially as \(3x^2 + 6x - 9 = 0\). Identifying and rewriting an equation in this standard format is the first step in solving quadratic equations effectively. Factoring is one common and powerful method for finding the roots or solutions of quadratic equations. Through factoring, expressions are simplified into product form, making it possible to find values that satisfy the equation.

Solving Equations

Solving equations is a fundamental skill in algebra that involves finding the value of the variable that makes the equation true. For the quadratic equation \(3x^2 + 6x - 9 = 0\), solving it involves multiple steps:

• Simplifying and rearranging the equation to standard form.
• Factoring or applying appropriate methods such as completing the square or using the quadratic formula.
• Setting each factor to zero to find the possible values for \(x\).

In the provided solution, factoring was used where the expression was rewritten as \((x + 3)(3x - 3) = 0\). This step interrelates with solving linear equations, where each bracket, representing a linear equation, is set to zero, yielding the individual solutions \(x = -3\) and \(x = 1\).

Breaking down equations into simpler parts makes the path to the solution clear and accessible.

Polynomials

Polynomials are algebraic expressions that consist of terms which are composed of variables and coefficients. They can be represented by the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\).

In the context of our exercise, the quadratic polynomial \(3x^2 + 6x - 9\) was central to the factorization method. Understanding how to manipulate polynomials is crucial as they form the building blocks of more complex algebraic expressions and equations.

When working with polynomials:

• Identify each term and its coefficient.
• Understand operations like addition, subtraction, multiplication, and division among polynomials.
• Apply distributive, associative, and commutative properties to simplify expressions.

Grasping these concepts aids significantly in breaking down and solving polynomial equations.

Algebraic Expressions

Algebraic expressions, such as \(3x^2 + 6x - 9\), are combinations of variables, constants, and operational symbols (+, -, *, /). They do not have an equality sign, unlike equations. Understanding these expressions is vital as they serve various functions in solving equations and modeling real-world scenarios.

Here's how algebraic expressions played a role in our solution:

• The expression on the right-hand side was expanded and simplified to obtain a quadratic form.
• Terms were collected to consolidate similar elements and facilitate easier manipulation.
• Factorization was performed to break down the expression into simpler parts, highlighting factors of the polynomial.

Mastering algebraic expressions is indispensable in building a strong foundation for tackling equations of varying complexities. Recognizing how to handle these expressions allows for systematic and efficient problem-solving in algebra.