Problem 48 - Algebra and Trigonometry

Question

Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) \(P(x)=x^{3}-x+1, \quad c=\frac{1}{4}\)

Step-by-Step Solution

Answer

\(P\left(\frac{1}{4}\right) = \frac{49}{64}\).

Step 1: Set up synthetic division

The polynomial is given by \( P(x) = x^3 - x + 1 \) and we want to evaluate \( P\left(\frac{1}{4}\right) \) using synthetic division. First, we must write the coefficients of the polynomial: \([1, 0, -1, 1]\). Notice that there is no \(x^2\) term, so we denote its coefficient as 0. Place \(\frac{1}{4}\), the value of \(c\), to the left of the synthetic division tableau. Then, write the coefficients of the polynomial horizontally to the right.

Step 2: Perform synthetic division

Begin the synthetic division by bringing down the leading coefficient, 1, unchanged. Multiply this number by \(\frac{1}{4}\) and write the result beneath the next coefficient. Add the coefficient and the product together and write the sum directly below. Repeat this process for all coefficients. The results should be arranged as follows:- The first number is the coefficient of \(x^2\).- The second number is for \(x\).- The third number represents the constant term.- The last number is the remainder.

Step 3: Calculate products and sums

1. Bring down the 1 (from \(x^3\)).2. Multiply 1 by \(\frac{1}{4}\) getting \(\frac{1}{4}\) and add to 0 (for \(x^2\)), resulting in \(\frac{1}{4}\).3. Multiply \(\frac{1}{4}\) by \(\frac{1}{4}\) getting \(\frac{1}{16}\) and add to -1 (\(x\)), providing \(-\frac{15}{16}\).4. Multiply \(-\frac{15}{16}\) by \(\frac{1}{4}\) to get \(-\frac{15}{64}\) and add to 1 (constant), resulting in \(\frac{49}{64}\). The remainder is \(\frac{49}{64}\), representing \(P\left(\frac{1}{4}\right)\).

Step 4: Apply the Remainder Theorem

According to the Remainder Theorem, when a polynomial \( P(x) \) is divided by \( x-c \), the remainder of that division is \( P(c) \). From the synthetic division process, we computed a remainder of \(\frac{49}{64}\). Therefore, \(P\left(\frac{1}{4}\right) = \frac{49}{64}\).

Key Concepts

Remainder Theorem

The Remainder Theorem is a powerful tool in polynomial mathematics. This theorem states that for any polynomial \( P(x) \), if you divide it by a linear divisor of the form \( x - c \), the remainder of this division is simply \( P(c) \). This theorem provides a quick method to find the value of the polynomial at any given point, \( c \), without needing to fully evaluate the polynomial expression directly.
Using this theorem is especially useful for checking potential roots of the polynomial or predicting polynomial behavior at specific values.

• This means you can avoid complex calculations by employing a division technique instead.
• It directly connects the division process and evaluation of polynomial expressions, making calculations more efficient.

With the Remainder Theorem, understanding the relationship between divisors and remainders in polynomials is greatly simplified, which is essential for both theoretical and practical applications in algebra.

Polynomial Evaluation

Polynomial evaluation involves calculating the specific value of a polynomial for a certain input or variable, often denoted as \( c \). When we evaluate a polynomial, we essentially substitute the variable, \( x \), with a numerical value to find the result of the polynomial expression at that point.
In the exercise, we are tasked with evaluating \( P(c) \) using \( c = \frac{1}{4} \) for the polynomial \( P(x) = x^3 - x + 1 \). This requires calculating the expression by either direct substitution or, as demonstrated in synthetic division, using the Remainder Theorem to simplify calculations.

• Direct substitution involves replacing every instance of \( x \) in the polynomial with \( \frac{1}{4} \).
• Using synthetic division simplifies finding \( P(c) \) and helps avoid extensive arithmetic calculations.

By understanding polynomial evaluation, you gain insights into the behavior of polynomial equations at specific values of \( x \). It is a foundational concept in algebra, crucial for solving equations and understanding polynomial functions.

Polynomial Division

Polynomial division is a method used to divide one polynomial by another. Although it may sound complicated, it's quite logical once you break it down into steps. In synthetic division, which is a simplified form of polynomial division, you can evaluate polynomials efficiently.
Synthetic division is especially useful when dividing a polynomial by a linear factor where the leading coefficient is 1. In our case, we used \( x - \frac{1}{4} \) as the divisor in synthetic division.

• Start by writing down the coefficients of the polynomial you want to divide.
• Set up a synthetic division tableau with \( c \) on the left and the coefficients on the right.
• Carry down the leading coefficient, then multiply and add across the tableau.

This method simplifies the division process by reducing many of the steps involved in regular division of polynomials. It's particularly useful for quickly finding remainders, as seen in our exercise, where the remainder directly equals \( P\left(\frac{1}{4}\right) \). Understanding synthetic division is very useful in solving polynomial equations and is a useful strategy beyond traditional algebraic methods.