Problem 39 - A Graphical Approach to Precalculus with Limits

Question

Use synthetic substitution to find \(P(k).\) $$k=0.5 ; \quad P(x)=x^{3}-x+4$$

Step-by-Step Solution

Answer

The value of \(P(0.5)\) is 3.625.

Step 1: Organize the Polynomial Coefficients

Start by identifying the coefficients of the polynomial \(P(x) = x^3 - x + 4\). In this polynomial, the coefficient of \(x^3\) is 1, the coefficient of \(x^2\) is 0 (since the term is missing), the coefficient of \(x\) is -1, and the constant term is 4. Thus, the coefficients are: 1, 0, -1, 4.

Step 2: Write the Synthetic Substitution Setup

Write down 0.5, the value of \(k\), followed by the coefficients of the polynomial below the horizontal line. This is the setup for synthetic substitution.\[\begin{array}{c|cccc}0.5 & 1 & 0 & -1 & 4 \ & & & & \\end{array}\]

Step 3: Perform the Synthetic Substitution

Bring down the leading coefficient, which is 1. Multiply it by 0.5 (the value of \(k\)), and write the result under the next coefficient.Add vertically to get the next value.Continue this process:1. Bring down 1.2. Multiply 1 by 0.5 to get 0.5, and add to 0 to get 0.5.3. Multiply 0.5 by 0.5 to get 0.25, and add to -1 to get -0.75.4. Multiply -0.75 by 0.5 to get -0.375, and add to 4 to get 3.625.

Step 4: Interpret the Results

The final number at the bottom, 3.625, is the value of \(P(0.5)\). This is because the last number in the synthetic substitution process represents \(P(k)\) when all operations are completed.

Key Concepts

Polynomial Coefficients

Coefficients play a crucial role in understanding polynomials. They are the numerical factors in each term of a polynomial expression. For instance, in the polynomial \(P(x) = x^3 - x + 4\), each term can be identified along with its coefficient:

• The coefficient of \(x^3\) is 1. This is because the term is simply \(x^3\), suggesting that it is multiplied by 1.
• For \(x^2\), there is no term explicitly showing, which means its coefficient is 0. This is an important detail since it indicates that \(x^2\) does not contribute to the polynomial's value.
• The coefficient of \(x\) is -1, given the term is -\(x\).
• The constant term, 4, stands alone as a number without a variable.

These coefficients together form a set: [1, 0, -1, 4]. These values are essential when using methods like synthetic division to evaluate or factorize polynomials. They guide us through the calculation process by showing how each term contributes to the overall expression.

Synthetic Division

Synthetic division is a streamlined method to divide polynomials and is particularly useful for dividing by linear terms of the form \(x - k\). Unlike long division, synthetic division simplifies the steps and works directly with the coefficients.

• Step 1: You start with the value of \(k\). In our exercise, \(k = 0.5\).
• Step 2: Write the coefficients of the polynomial in a row: [1, 0, -1, 4]. These are positioned next to a vertical line with \(k\) outside the line.
• Step 3: Begin the process: Bring down the leading coefficient (1 in this case) to the row below the line.

Now, we proceed by iteratively multiplying the last number in the row below by \(k\) and adding it to the next coefficient in the original list. This simplicity can make complex polynomial division more approachable and reduces errors.
The final result on the bottom row gives important insights, either revealing the remainder when dividing or providing the evaluated polynomial value, which in this exercise, is \(P(0.5)\).

Polynomial Evaluation

Polynomial evaluation involves calculating the value of a polynomial expression when substituting a specific number for the variable \(x\). Synthetic substitution is a technique closely related to synthetic division and is particularly efficient for evaluating polynomials at a given value of \(x\) quickly.
For our exercise, synthetic substitution was performed to find \(P(0.5)\). Through this process:

• Each step combines multiplication and addition of coefficients with the value of \(k\) (0.5 here), moving through each term's coefficient.
• The last number obtained represents the result of the polynomial evaluation at \(x = 0.5\).

This process concludes with \(3.625\) as the evaluated result. Understanding synthetic substitution as a form of polynomial evaluation offers a powerful tool for handling polynomial functions, particularly those of higher degree with greater complexity.