Problem 40 - Algebra and Trigonometry

Question

Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) \(P(x)=x^{3}-x^{2}+x+5, \quad c=-1\)

Step-by-Step Solution

Answer

The value of \(P(-1)\) is 6.

Step 1: Set up synthetic division

Start by setting up the synthetic division. Write the coefficients of the polynomial \(P(x)=x^3-x^2+x+5\) in descending order of power: \([1, -1, 1, 5]\). The divisor is \(c = -1\). Place \(-1\) to the left of the coefficients.

Step 2: Bring down the leading coefficient

Take the leading coefficient (the first number in the list, which is \(1\)) and bring it down below the line. This is your starting point for synthetic division.

Step 3: Multiply and add iteratively

Multiply \(-1\) (the number on the left) by the number just brought down (\(1\)). Write \(-1\) under the next coefficient \(-1\) and add them: \(-1) + (-1) = -2\). Repeat this process: multiply \(-1\) by \(-2\), place the result under the next coefficient \(1\), and add them: \(1 + (-2) = -1\). Finally, multiply \(-1\) by \(-1\), put the result under \(5\), and add them: \(5 + 1 = 6\).

Step 4: Interpret the result

The bottom row now represents the coefficients of the quotient polynomial and the remainder: \([1, -2, -1, 6]\). The remainder is the last number, \(6\). According to the Remainder Theorem, \(P(-1)\) is equal to this remainder.

Key Concepts

Polynomial Division

Polynomial division is a technique used to divide one polynomial by another, simpler polynomial. It is similar to long division with numbers, but involves variables and coefficients. In this context, we will specifically refer to synthetic division, a shortcut that simplifies the division process when dealing with polynomials. Synthetic division can only be used when dividing by a linear polynomial, often represented in the form of \(x - c\). In this exercise, our polynomial is \(P(x) = x^3 - x^2 + x + 5\) and we are dividing it by \(x + 1\) (or equivalently, \(c = -1\)). This makes synthetic division a perfect tool for our needs as it:

• Saves time and reduces the complexity of calculations
• Requires only the coefficients of the polynomial
• Focuses on multiplication and addition operations

The advantage of synthetic division is that it is quick and can easily be calculated by hand, especially with practice. It involves setting up a grid that allows us to handle the coefficients directly and compute the result step by step. This method is wonderfully structured to leave us with the quotient of our division and a remainder that tells us how well our divisor fits into the original polynomial.

Remainder Theorem

The Remainder Theorem is a straightforward yet powerful tool in polynomial algebra. It provides a quick way to evaluate the remainder of a polynomial division without performing the complete division process.

The theorem states that when a polynomial \(P(x)\) is divided by a binomial of the form \(x - c\), the remainder of the division is equal to \(P(c)\). This means that by simply substituting \(c\) into the polynomial, we can determine the remainder without having to perform any actual division.

• Using our example, \(P(-1)\) is the remainder when \(P(x)\) is divided by \(x + 1\).
• In synthetic division, this remainder is the final number in our row of computed values.

The Remainder Theorem is particularly useful for quickly checking results and verifying roots of polynomials. If \(P(c) = 0\), then \(x - c\) is a factor of the polynomial. This understanding simplifies many polynomial-related problems and is a handy shortcut.

Polynomial Evaluation

Evaluating a polynomial simply means finding the value of a polynomial function for a given input. This task involves substituting a value for the variable \(x\) and performing the arithmetic operations to solve for the polynomial's value.

In the context of our example, polynomial evaluation is needed to confirm the remainder we obtain from synthetic division. By substituting \(c = -1\) directly into \(P(x) = x^3 - x^2 + x + 5\), the evaluation becomes a simple matter of solving:

• \((-1)^3 - (-1)^2 + (-1) + 5\)
• = -1 - 1 - 1 + 5
• = 2

This result verifies our synthetic division calculation, as both methods yield a remainder of 6, demonstrating that our process was correct. Evaluating polynomials in this way is a fundamental skill in algebra, as it ensures that calculations related to polynomials are both accurate and reliable.