Problem 103 - Elementary and Intermediate Algebra

Question

Complete each table. See Example 11. $$ \begin{array}{|c|c|} \hline x & {(x+1)(x+5)} \\ \hline-1 & {} \\ \hline-5 & {} \\ \hline-6 & {} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

For \(x = -1\), value is 0; for \(x = -5\), value is 0; for \(x = -6\), value is 5.

Step 1: Understanding the Expression

The expression given in the table is \((x+1)(x+5)\). This is a factored quadratic expression that we need to evaluate for different values of \(x\).

Step 2: Calculate for \(x = -1\)

Substitute \(-1\) into the expression: \((-1 + 1)(-1 + 5)\). This simplifies to \((0)(4) = 0\). Thus, when \(x = -1\), the value is 0.

Step 3: Calculate for \(x = -5\)

Substitute \(-5\) into the expression: \((-5 + 1)(-5 + 5)\). This simplifies to \((-4)(0) = 0\). Hence, when \(x = -5\), the value is 0.

Step 4: Calculate for \(x = -6\)

Substitute \(-6\) into the expression: \((-6 + 1)(-6 + 5)\). This simplifies to \((-5)(-1) = 5\). Therefore, when \(x = -6\), the value is 5.

Key Concepts

Factored Form

Understanding quadratic expressions is key in algebra, and the factored form plays a crucial role. A quadratic expression in factored form is written as \((x + a)(x + b)\), where \(a\) and \(b\) are numbers. This form is particularly useful for solving equations as it simplifies the process of finding the roots.In the factored form, each part of the expression represents a line intercept on a graph when the expression is set to zero. For instance, in our example, \((x+1)(x+5)\), the roots of the quadratic equation \(x+1=0\) and \(x+5=0\) are \(x=-1\) and \(x=-5\) respectively. This means when graphed, the curve will cross the \(x\)-axis at these two points.This form is beneficial when performing operations like evaluating expressions, as it gives a clear way to substitute specific values of \(x\) directly into the expression.

Substitution

Substitution is a fundamental technique in algebra, used to evaluate expressions at specific values of variables. To apply substitution in the context of a quadratic expression, you'll replace \(x\) with the given number and simplify step by step.For example, to substitute \(x=-1\) in \((x+1)(x+5)\), follow these steps:

• Replace \(x\) with \(-1\) producing \((-1+1)\) and \((-1+5)\).
• Simplify both parts to get \(0\) and \(4\).
• Multiply the results: \(0 \times 4 = 0\).

Substitution not only helps to evaluate the expression but also ties into solving equations and graphing purposes. Doing these calculations manually for a range of \(x\) values can give insight into the behavior of the quadratic expression.

Evaluation of Algebraic Expressions

The process of evaluating algebraic expressions involves calculating their values by substituting variables with numbers. This operation is crucial to understanding how expressions behave under different conditions.Consider \((x+1)(x+5)\). To evaluate this expression:

• Select a value for \(x\).
• Substitute this value in place of \(x\) in the expression.
• Carry out the arithmetic operations to get the final result.

The key here is to follow the order of operations: parentheses first, then multiplication and addition/subtraction. For example, when \(x=-6\), substituting yields \((-6+1)(-6+5)\), which simplifies to \((-5)(-1)\) or \(5\). This step-by-step approach ensures accuracy and helps visualize the impact of different values in the expression. This kind of evaluation becomes particularly powerful when plotting graphs or solving real-life problems.