Problem 17 - Algebra 2

Question

Use synthetic substitution to find \(g(3)\) and \(g(-4)\) for each function. $$ g(x)=x^{6}-4 x^{4}+3 x^{2}-10 $$

Step-by-Step Solution

Answer

g(3) = 422 and g(-4) = 3110.

Step 1: Set up synthetic substitution for g(3)

To find \(g(3)\), use the coefficients of the polynomial \(g(x) = x^6 - 4x^4 + 3x^2 - 10\). Arrange them: \([1, 0, -4, 0, 3, 0, -10]\). Note placeholders (zeroes) for missing terms \(x^5\), \(x^3\), and \(x\).

Step 2: Perform synthetic substitution for g(3)

Place 3 outside the synthetic division. Bring down the first coefficient (1). Multiply by 3 and add to the next coefficient (0), repeat. Resulting operations in order: 1. \(1 \times 3 + 0 = 3\)2. \(3 \times 3 + (-4) = 5\)3. \(5 \times 3 + 0 = 15\)4. \(15 \times 3 + 3 = 48\)5. \(48 \times 3 + 0 = 144\)6. \(144 \times 3 + (-10) = 422\). Final result: \(g(3) = 422\).

Step 3: Set up synthetic substitution for g(-4)

Repeat the setup for \(g(-4)\) using the same coefficients \([1, 0, -4, 0, 3, 0, -10]\). Place -4 outside.

Step 4: Perform synthetic substitution for g(-4)

Place -4 outside the synthetic division. Bring down the first coefficient (1). Multiply by -4 and add to the next coefficient (0), repeat. Resulting operations in order: 1. \(1 \times (-4) + 0 = -4\)2. \(-4 \times (-4) + (-4) = 12\)3. \(12 \times (-4) + 0 = -48\)4. \(-48 \times (-4) + 3 = 195\)5. \(195 \times (-4) + 0 = -780\)6. \(-780 \times (-4) + (-10) = 3110\). Final result: \(g(-4) = 3110\).

Key Concepts

Polynomials

A polynomial is a mathematical expression involving a sum of powers of a variable, each multiplied by a coefficient. These expressions are commonly used because they can represent a wide variety of functions. An example of a polynomial is the function given in the exercise:

• Format: It's expressed as a sum like: \(x^6 - 4x^4 + 3x^2 - 10\).
• Degrees: The degree of a polynomial is the highest power of the variable in the expression, which in this case is 6 due to the \(x^6\) term.
• Coefficients: These are the numbers multiplying each power of the variable; in our polynomial, they are 1, -4, 3, and -10 for the respective powers of x.
• Variables: The variable here is \(x\), and it can take on various values.

Polynomials are foundational in algebra and appear in various fields of mathematics. They are used to model and solve problems ranging from simple algebraic equations to complex calculus problems.

Synthetic Division

Synthetic division is a simplified way of dividing a polynomial by another polynomial of the form \(x - c\). It's much quicker and less complex than long division, which makes it a popular tool in algebra. Here's how it works with the given polynomial in the exercise:

• Setup: Arrange the coefficients of the polynomial in descending order of their respective powers, filling in zeros for any missing terms, such as \(x^5, x^3, \) and \(x\).
• Order: When you perform synthetic division, you only deal with the coefficients (no variable shown), which results from repetitive multiplication and addition.
• Procedure: Given a number \(c\) (like 3 or -4 from our exercise), it's placed outside and used to perform operations starting from the leading coefficient.
• Benefits: Synthetic division not only helps in finding values (like \(g(3)\)) quickly but also reduces errors often associated with more complex arithmetic.

The efficiency makes it ideal for evaluating polynomial functions at specific values without getting bogged down by intricate algebra.

Evaluating Polynomial Functions

Evaluating polynomial functions involves calculating the output (or result) of a polynomial for given input values. It allows you to understand how the polynomial behaves or changes. Here's how it works using synthetic substitution:

• Alexander:"): Given a polynomial \(g(x)\), you find \(g(c)\) by substituting \(c\) into the polynomial. In our exercise, it was done for \(x = 3\) and \(x = -4\).
• Simplicity of Synthetic Substitution: This method uses synthetic division to simplify repetitive calculations. It involves multiplying and adding coefficients rather than substituting each \(c\) value directly into the polynomial.
• Benefits: Evaluating allows you to determine specific points of the polynomial’s graph, helping in sketching it or solving for algebraic solutions. It also enables you to verify roots and intercepts efficiently.

Evaluating helps in data analysis, physics simulations, and even economics for functions that change over time or scenarios, making it a powerful tool in diverse applications.