Problem 40 - Algebra and Trigonometry

Question

\(39-51\) . Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) $$ P(x)=2 x^{2}+9 x+1, \quad c=\frac{1}{2} $$

Step-by-Step Solution

Answer

\(P\left(\frac{1}{2}\right) = 6\).

Step 1: Identify Coefficients

Identify the coefficients of the polynomial \(P(x) = 2x^2 + 9x + 1\). The coefficients are 2, 9, and 1.

Step 2: Setup Synthetic Division

For synthetic division to evaluate \(P(c)\), set \(c = \frac{1}{2}\). Our division process begins with the first coefficient, 2.

Step 3: First Row of Synthetic Division

Write the first coefficient (2) in the result row. Multiply it by \(\frac{1}{2}\), and write that product (which is 1) under the next coefficient (9).

Step 4: Add Down and Multiply

Add 9 and 1 to get 10. Multiply 10 by \(\frac{1}{2}\) to get 5. Write 5 under the next coefficient (1).

Step 5: Final Addition

Add 1 and 5 to get 6. This is the remainder, which is \(P\left(\frac{1}{2}\right)\). The remainder theorem confirms that 6 is the evaluation of \(P(x)\) at \(x = \frac{1}{2}\).

Key Concepts

Polynomial Evaluation

Polynomial evaluation is the process of finding the value of a polynomial at a given point. It's like asking, "What is the result when we plug this specific number into the polynomial function?" In our exercise, we aim to evaluate the polynomial \(P(x) = 2x^2 + 9x + 1\) when \(x\) is \(\frac{1}{2}\).
This concept allows us to determine a specific output by substituting \(x\) with our desired value, \(c\). By using synthetic division, we efficiently handle this substitution and simplify the polynomial, ultimately finding the value of \(P(\frac{1}{2})\).
Synthetic division streamlines polynomial evaluation by systematically applying the polynomial's coefficients through an organized procedure, reinforcing the methodical nature of mathematics by producing accurate results swiftly.

Remainder Theorem

The Remainder Theorem is a key concept in polynomial division. It states that the remainder of the division of a polynomial \(P(x)\) by a linear divisor \(x - c\) is exactly \(P(c)\). This means that evaluating a polynomial at point \(c\) is equivalent to doing synthetic division and finding the remainder.
In the exercise, after performing synthetic division with \(c = \frac{1}{2}\), the remainder is found to be \(6\).
Happily, according to the Remainder Theorem, this remainder confirms the polynomial's evaluation at that specific point. It serves as an efficient shortcut, avoiding the labor-intensive work of direct substitution and enabling students to gauge their understanding quickly.

Polynomial Coefficients

Polynomial coefficients are the numerical factors that multiply each term of the polynomial. They play a crucial role in both synthetic division and polynomial evaluation. In our specific example \(P(x) = 2x^2 + 9x + 1\), the coefficients are 2, 9, and 1.
When employing synthetic division, these coefficients are manipulated, step by step, in a streamlined table setup that facilitates the calculation of \(P(c)\).
Understanding the role of coefficients enhances comprehension since they dictate the shape and behavior of a polynomial. During synthetic division, each coefficient is strategically used and modified until we confidently arrive at the evaluation, illuminating the dynamic nature of polynomials.