Problem 51 - Precalculus Mathematics for Calculus

Question

Use synthetic division and the Remainder Theorem to evaluate \(P(c)\). $$P(x)=x^{3}+2 x^{2}-3 x-8, \quad c=0.1$$

Step-by-Step Solution

Answer

Using synthetic division, \(P(0.1) = -8.279\).

Step 1: Setup Synthetic Division

First, we are given the polynomial \(P(x) = x^3 + 2x^2 - 3x - 8\) and a value \(c = 0.1\). We arrange the coefficients of \(P(x)\), which are \([1, 2, -3, -8]\), in a row, and place \(c = 0.1\) outside the division framework.

Step 2: Start Synthetic Division

Begin by bringing down the first coefficient \(1\) as it is. This value becomes the first row's entry directly below the line.

Step 3: Multiply and Add

Multiply \(0.1\) (our \(c\)) by the value below the line (\(1\)), which gives \(0.1\). Add this to the next coefficient \(2\) to get \(2.1\). Write \(2.1\) below the line next to the result from Step 2.

Step 4: Repeat Process

Continue by multiplying \(2.1\) by \(0.1\), resulting in \(0.21\). Add \(0.21\) to the next coefficient \(-3\), yielding \(-2.79\). Place \(-2.79\) below the line.

Step 5: Final Multiplication and Addition

Multiply \(-2.79\) by \(0.1\) to get \(-0.279\). Add \(-0.279\) to the last coefficient \(-8\), resulting in \(-8.279\). Record \(-8.279\) as the remainder.

Step 6: Interpret the Remainder

The Remainder Theorem tells us that the remainder from this synthetic division, \(-8.279\), is the value of \(P(c)\) when \(c = 0.1\). Therefore, \(P(0.1) = -8.279\).

Key Concepts

Remainder Theorem

The Remainder Theorem is a simple yet powerful tool in algebra. It states that if you divide a polynomial \(P(x)\) by a binomial of the form \(x - c\), the remainder of this division is equal to \(P(c)\). This means that instead of fully dividing a polynomial, you can use the Remainder Theorem to quickly find out the value of the polynomial at \(c\).
This theorem becomes incredibly handy when you need to evaluate polynomials at specific points, as it saves time and effort. For instance, completing a synthetic division tells you the remainder, which is the same as \(P(c)\). Understanding this connection helps identify and calculate polynomial values efficiently.

Polynomial Evaluation

Evaluating a polynomial is the process of calculating the value of the polynomial for a specific variable input. Using the polynomial \(P(x) = x^3 + 2x^2 - 3x - 8\) as an example, if we wish to find \(P(0.1)\), we could substitute \(x = 0.1\) into each term and solve. However, this can be tedious for more complex or higher-degree polynomials.
Synthetic division, combined with the Remainder Theorem, offers a streamlined method for evaluation. By performing synthetic division with \(c = 0.1\), you obtain the remainder, which directly corresponds to \(P(0.1)\). This method is efficient and reduces errors associated with manual computation.

Division of Polynomials

Division of polynomials can be cumbersome when performed manually through traditional long division methods. Luckily, synthetic division provides a simplified technique especially suited for dividing polynomials by linear expressions of the form \(x - c\).
The process involves setting up a synthetic division framework using the coefficients of the polynomial and repeatedly multiplying and adding with the value selected for \(c\). This streamlined setup can save time and effort in computations. The result will not only reveal the remainder but also assist in finding factors of the polynomial or simplifying polynomial expressions.

Polynomial Remainder

The polynomial remainder is the outcome left after dividing one polynomial by another. In the context of synthetic division, this remainder provides critical information. Notably, as per the Remainder Theorem, it equates to the value of the polynomial at the divisor \(c\).
For the given example, after performing synthetic division on \(P(x) = x^3 + 2x^2 - 3x - 8\) with \(c = 0.1\), the remainder was found to be \(-8.279\). Therefore, \(P(0.1) = -8.279\). This illustrates how useful the remainder is in polynomial calculations and evaluations.