Problem 14 - Precalculus with Calculus Previews

Question

Proceed as in Example 2 and use the Remainder Theorem to find \(r\) when \(f(x)\) is divided by the given linear polynomial. $$ f(x)=5 x^{3}+x^{2}-4 x-6 ; x+1 $$

Step-by-Step Solution

Answer

The remainder \(r\) is \(-6\).

Step 1: Identify the Linear Polynomial

The given linear polynomial is \(x + 1\). We need to find the remainder \(r\) when \(f(x)\) is divided by this polynomial.

Step 2: Use the Remainder Theorem

The Remainder Theorem states that the remainder of the division of \(f(x)\) by \(x-a\) is \(f(a)\). Here, \(a = -1\) because the polynomial is \(x + 1 = x - (-1)\).

Step 3: Substitute \(x = -1\) into \(f(x)\)

Calculate \(f(-1)\) by substituting \(x = -1\) into the polynomial: \[ f(-1) = 5(-1)^3 + (-1)^2 - 4(-1) - 6 \] Simplify this expression to find the remainder.

Step 4: Simplify the Expression

Compute each term: \(-5 + 1 + 4 - 6 = -6\). Thus, \(f(-1) = -6\).

Step 5: State the Remainder

The remainder \(r\) when \(f(x)\) is divided by \(x+1\) is \(-6\).

Key Concepts

Polynomials

Polynomials are expressions consisting of variables and coefficients, organized in a way where the powers of the variable are whole numbers. In simpler terms, they are algebraic expressions with terms like \(5x^3\), \(x^2\), and so on. The degree of the polynomial is determined by the highest power of the variable present in the expression. For example, in \(f(x)=5x^3 + x^2 - 4x - 6\), the highest power of \(x\) is 3, making it a cubic polynomial.

• Constant Term: The term that does not contain any variable, such as \(-6\) in this case.
• Coefficient: The numerical factor in each term. For instance, 5 is the coefficient of \(x^3\).

Understanding the structure of polynomials is crucial when working with mathematical operations such as addition, subtraction, multiplication, and division involving them. The Remainder Theorem, as applied in the given exercise, assists in determining the remainder when a polynomial is divided by a linear polynomial.

Division of Polynomials

The division of polynomials is similar to the division of integers but follows a different set of rules to determine quotient and remainder. When dividing polynomials by linear polynomials, the Remainder Theorem provides a quick way to find the remainder without performing complete division. Let's break down how this procedure works:

• Dividend: The polynomial being divided. In the example, it is \(f(x)=5x^3 + x^2 - 4x - 6\).
• Divisor: The polynomial by which we are dividing. In the example, the divisor is \(x+1\).
• Remainder: The leftover part after division. What you solve for using the Remainder Theorem in this problem.

The Remainder Theorem specifically states that for any polynomial \(f(x)\), if you divide it by \(x-a\), the remainder of the division is \(f(a)\). This theorem simplifies the process by allowing us to evaluate the polynomial at a specific point (\(x = a\)) rather than performing long division.

Evaluation of Functions

Evaluation of functions is the process of calculating the value of a function for a specific input. It involves substituting the input value into the function equation to determine the result. For instance, in our exercise with \(f(x) = 5x^3 + x^2 - 4x - 6\), evaluating the function at \(x = -1\) required inserting \(-1\) wherever \(x\) appeared:

\[ f(-1) = 5(-1)^3 + (-1)^2 - 4(-1) - 6\]

• Calculate each term individually: \(-5, 1, 4, -6\).
• Add them together to get the function's value for that input: \(-5 + 1 + 4 - 6 = -6\).

This process is straightforward but requires careful computation to ensure accuracy, especially in polynomials with multiple terms. The result, \(-6\), shows the remainder when the polynomial is divided by \(x + 1\), illustrating the practical use of function evaluation with the Remainder Theorem.