Problem 10 - Precalculus Mathematics for Calculus

Question

Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express the quotient \(P(x) / D(x)\) in the form $$\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$$ $$P(x)=x^{3}+6 x+5, \quad D(x)=x-4$$

Step-by-Step Solution

Answer

The quotient is \(x^2 + 4x + 22\) with a remainder of 93.

Step 1: Identify the method

We are given polynomials \(P(x) = x^3 + 6x + 5\) and \(D(x) = x - 4\). Since \(D(x)\) is a linear polynomial, synthetic division is suitable for this case.

Step 2: Set up synthetic division

Write down the coefficients of \(P(x)\): 1 (for \(x^3\)), 0 (for \(x^2\)), 6 (for \(x\)), and 5. The divisor \(D(x) = x - 4\) means we will use \(4\) as the synthetic divisor.

Step 3: Perform synthetic division

Start by bringing the first coefficient (\(1\)) down. Multiply it by the divisor (\(4\)), and write the result under the next coefficient (\(0\)). Add and continue this process:- 1: Bring down the 1.- 4: Multiply 4 by 1 and add to 0 (gives 4).- 24: Multiply 4 by 4 and add to 6 (gives 22).- 88: Multiply 4 by 22 and add to 5 (gives 93).

Step 4: Write the result

The quotient is obtained from the results: \(x^2+4x+22\). The remainder is 93. Hence, the expression for \(\frac{P(x)}{D(x)}\) is \(x^2 + 4x + 22 + \frac{93}{x-4}\).

Key Concepts

Synthetic Division

Synthetic Division is a method used to divide a polynomial by a linear divisor of the form \(x - c\). It is a simplified version of long division, focusing only on the coefficients of the polynomial, making the process quicker and more efficient. For our example, let's break down how synthetic division works.

To start with synthetic division, first, identify the coefficients of the polynomial you wish to divide. For \(P(x) = x^3 + 6x + 5\), the coefficients are:

• 1 for \(x^3\)
• 0 for \(x^2\) since it is missing
• 6 for \(x\)
• 5 as the constant term

The linear divisor \(D(x) = x - 4\) suggests we use \(c = 4\) in our synthetic division.

The process involves the following steps:

• Bring down the first coefficient (1) as it is.
• Multiply the value brought down by \(c\) (which is 4 in this case).
• Add the result to the next coefficient and repeat the process until completion.

Synthetic division yields the quotient and can handle polynomial division swiftly, as shown where we obtained \(x^2 + 4x + 22\) with a remainder of 93.

Quotient and Remainder

The concepts of Quotient and Remainder are fundamental in polynomial division. When you divide any polynomial \(P(x)\) by another polynomial \(D(x)\), you typically end up with a quotient \(Q(x)\) and a remainder \(R(x)\). The equation can be expressed as:

\[\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}\]

In our example, the division yielded a quotient \(Q(x)\) of \(x^2 + 4x + 22\). This was obtained through the synthetic division process. The remainder when \(P(x)\) was divided by \(D(x)\) is 93.

• The quotient \(Q(x)\) represents the portion of \(P(x)\) that divides evenly by \(D(x)\).
• The remainder \(R(x)\) is what is left over after division and must be of a smaller degree than \(D(x)\).

This format is essential for organizing polynomials after division, useful in various applications including simplifications in algebra.

Rational Expression

A Rational Expression is a fraction where both the numerator and the denominator are polynomials. The expression \(\frac{P(x)}{D(x)}\) can often be rewritten in terms useful for various computations and applications.

In our specific division problem, upon dividing \(P(x) = x^3 + 6x + 5\) by \(D(x) = x - 4\), we expressed the rational expression as a sum of a polynomial and a proper fraction:
\[x^2 + 4x + 22 + \frac{93}{x-4}\].
Rational expressions can usually simplify and help in solving equations or optimizing in calculus. Key aspects include focusing on:

• Simplifying expressions by factoring.
• Performing operations like addition, subtraction, multiplication, and division on rational expressions, following algebraic rules.
• Considering the domain—the set of all possible values—which might exclude any value that makes the denominator zero.

Working with rational expressions involves ensuring simplicity and clarity, making them manageable and useful tools in both basic and advanced mathematics.