Problem 73 - Elementary Algebra

Question

Solve. $$ 6(x+1)(x-1)=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = -1\) and \(x = 1\).

Step 1: Understand the Equation

The given equation is \(6(x+1)(x-1)=0\). This is a product of three factors: \(6\), \((x+1)\), and \((x-1)\). According to the Zero Product Property, if a product of several factors is equal to zero, then at least one of the factors must be zero.

Step 2: Identify Factors Involved

We observe that the equation consists of the constant factor \(6\) and two linear binomial factors: \((x+1)\) and \((x-1)\). The constant 6 cannot be zero, so we will focus on the binomial factors.

Step 3: Apply Zero Product Property

Set each of the binomial factors to zero, since that's necessary to satisfy the Zero Product Property. This means we write two separate equations: \(x+1=0\) and \(x-1=0\).

Step 4: Solve Each Equation for x

Solve each equation separately:For \(x+1=0\):\[x = -1\]For \(x-1=0\):\[x = 1\]

Step 5: Provide the Solution Set

Since both equations derived from the Zero Product Property hold true separately, the solution set for the original equation is \(x = -1\) and \(x = 1\).

Key Concepts

Understanding Binomial Factors

In algebra, a binomial factor is an expression that contains two terms joined by either addition or subtraction. In the equation \(6(x+1)(x-1)=0\), we have two binomial factors: \((x+1)\) and \((x-1)\). These are called binomial factors because each has two parts:

• \(x\) is a variable, representing an unknown value.
• The constants \(1\) and \(-1\) add or subtract from \(x\).

The primary function of these factors is to show how they interact to form a product that equals zero when multiplied together with the constant 6. Notice that the constant 6 does not affect the solution because it cannot be zero. Only the binomial factors can potentially be zero. Understanding this helps us apply the Zero Product Property effectively.

The Process of Solving Equations

Solving equations is a fundamental process in mathematics that involves finding the value of a variable that makes the equation true. For the equation \(6(x+1)(x-1)=0\), this process starts by using the Zero Product Property. This property states that if the product of factors equals zero, at least one of the factors must be zero. Here’s how it works step-by-step:

• Identify all factors in the equation.
• Exclude any constant terms that cannot be zero, like the number 6 in this example.
• Set each binomial factor equal to zero separately.
• Solve each resulting equation to find the value of \(x\).

By treating each binomial factor as a potential solution separately, we can derive all possible values of \(x\) that satisfy the given equation.

Characteristics of Linear Equations

Linear equations are equations of the first degree, meaning that they form straight lines when graphed and have no exponents higher than one. In our example equation, each binomial factor \((x+1)\) and \((x-1)\) represents a linear equation on its own. When solved, linear equations have the following characteristics:

• They are simple and involve only one unknown variable, \(x\), raised to the power of one.
• Solving them involves basic arithmetic to isolate \(x\) on one side of the equation.
• The solutions are always single values (in this case, \(x = -1\) and \(x = 1\)) or none at all if no valid solutions exist.

Understanding linear equations is key to mastering more complex algebraic concepts, as they form the foundational skills required to tackle more complicated mathematical challenges.