Problem 69 - Algebra 2

Question

Use synthetic substitution to find \(f(-2)\) for \(f(x)=x^{3}+6 x-2\)

Step-by-Step Solution

Answer

\( f(-2) = -22 \)

Step 1: Identify the polynomial and value to substitute

The given polynomial is \( f(x) = x^3 + 6x - 2 \). The value we need to substitute is \( x = -2 \). We will use synthetic substitution to find \( f(-2) \).

Step 2: Set up synthetic substitution

Write down the coefficients of the polynomial. The polynomial \( f(x) = x^3 + 0x^2 + 6x - 2 \) has coefficients: 1 (for \(x^3\)), 0 (for \(x^2\)), 6 (for \(x\)), and -2 (constant). Set them up as a row: 1, 0, 6, -2.

Step 3: Start the synthetic division process

Bring down the first coefficient: 1. This becomes the start of your bottom row in synthetic division.

Step 4: Multiply and add

Multiply the root value (-2) by the number just written in the bottom row (1) and write the result (which is -2) in the next column of the second row. Add this to the second coefficient (0) in the top row, resulting in -2, which you write in the bottom row.

Step 5: Continue the synthetic division process

Multiply -2 (from bottom row) by -2 and place the result (4) in the next column of the second row. Add to the third coefficient (6), resulting in 10, which you write in the bottom row.

Step 6: Complete the synthetic division steps

Multiply 10 by -2 and write -20 in the next column of the second row. Add to the constant (-2), resulting in -22, which you write in the bottom row. This is \( f(-2) \).

Key Concepts

Understanding Polynomial Evaluation

Polynomial evaluation is a fundamental concept in algebra where you find the value of a polynomial function at a specific input value. This means calculating what happens when you "plug in" a particular number for "x" in your polynomial equation. For instance, if you're given a polynomial function like \( f(x) = x^3 + 6x - 2 \), and you want to find \( f(-2) \), you replace each "x" in the equation with -2. This results in the numerical value that the polynomial is equal to at \( x = -2 \).

Doing this manually involves substituting and then computing each term separately. However, using methods like synthetic substitution can streamline the process, especially for higher-degree polynomials. This method breaks the process into simpler arithmetic operations, making it much easier and quicker to evaluate.

Exploring Synthetic Division

Synthetic division is a simplified form of polynomial division, mainly used for dividing polynomials by linear factors of the form \( x - c \). It's faster than traditional long division and is highly useful when evaluating polynomials at specific points. In synthetic division, you only need the coefficients of the polynomial and the specific value for substitution.

Here's a brief overview of the steps:

• List the coefficients of the polynomial. If any term is missing, use 0 as its coefficient.
• The number you substitute (like -2 in the example) goes to the left outside the division bracket.
• The first coefficient drops down unchanged.
• Multiply that coefficient by the substitution value and add it to the next coefficient.
• Continue the process across all coefficients, always multiplying and then adding.

By the end of this process, the last value you get is the remainder, representing the value of the polynomial evaluated at \( x = c \). This makes synthetic division an efficient tool for polynomial evaluation.

The Role of Algebraic Functions

Algebraic functions are advanced functions in algebra built up from polynomial expressions. They serve as the backbone of various algebraic manipulations and solutions, including polynomial evaluation. An algebraic function can be as simple as a linear function like \( f(x) = 2x + 3 \), or as complex as a high-degree polynomial such as \( f(x) = x^5 - 4x^3 + x - 7 \).

Each polynomial, a primary example of algebraic functions, represents a sum of terms, each consisting of a constant multiplied by "x" raised to a non-negative integer power. These expressions can represent real-world scenarios, making algebraic functions a versatile tool for modeling and problem-solving.

In the context of synthetic substitution, you're often working with algebraic functions to easily find values at given inputs. Understanding the structure of these functions helps in handling complex algebraic operations with confidence.