Problem 36 - Intermediate Algebra

Question

Perform each division. \(\frac{s^{3}+10 s^{2}+17 s+12}{s+8}\)

Step-by-Step Solution

Answer

The result is \(s^2 + 2s + 1 + \frac{4}{s+8}\).

Step 1: Identify the Division Type

The given expression is a polynomial division problem. We are dividing the polynomial in the numerator, \(s^3+10s^2+17s+12\), by the linear polynomial \(s+8\). We'll use synthetic division as \(s+8\) is a linear binomial.

Step 2: Set Up Synthetic Division

For synthetic division with \(s+8\), we use \(-8\) (the opposite sign of 8) in the division process. List the coefficients of the polynomial: 1, 10, 17, and 12.

Step 3: Start the Division Process

Bring down the leading coefficient. The first coefficient \(1\) is brought down unchanged.

Step 4: Multiply and Add

Multiply \(-8\) by the leading coefficient (1) and add the result to the second coefficient (10). This gives us: \(-8 \times 1 + 10 = 2\).

Step 5: Repeat Multiply and Add

Continue the pattern: Multiply \(-8\) by the result from the previous step (2) and add it to the next coefficient (17): \(-8 \times 2 + 17 = 1\).

Step 6: Continue the Process

Repeat another round: Multiply \(-8\) by the latest result (1) and add it to the final coefficient (12): \(-8 \times 1 + 12 = 4\).

Step 7: Interpret the Results

The numbers obtained are the coefficients of the quotient and the remainder. The quotient is \(s^2 + 2s + 1\), and \(4\) is the remainder.

Key Concepts

Polynomial Division

Polynomial division is a fundamental concept in algebra, allowing us to divide polynomials similarly to how we divide numbers. With polynomial division, we're interested in breaking down a polynomial (the dividend) by another polynomial (the divisor) to find a quotient and possibly a remainder.
At its core, polynomial division can be compared to long division used with numbers. However, when dealing with polynomials, especially higher-degree ones, the process can become more complex.
The aim is to simplify equations and solve for unknowns, making sense of the mathematical relationships within the expression. In the problem given, we're dividing a cubic polynomial, which means it has three terms in descending order of power, by a linear polynomial to simplify the expression.

Linear Polynomials

A linear polynomial is one of the simplest forms of polynomials. It has either one variable raised to the first power, often shaped like \(ax + b\). Unlike polynomials with higher degrees, linear polynomials have a straight-line graph, hence the name 'linear.'
They serve as the basic building blocks of algebra and are easy to manipulate when solving equations or performing other operations. In the original exercise, the linear polynomial \(s+8\) is what we use to divide the cubic polynomial.
The simplicity of linear polynomials, in this case, allows us the use of synthetic division as opposed to the more tedious long division process. Recognizing the linear nature of the divisor helps streamline calculations when using synthetic division.

Synthetic Division Process

The synthetic division process is a shortcut that makes dividing polynomials more straightforward, especially when one of them is linear. This process is simpler than long division and involves fewer steps.
Here's a quick outline of how synthetic division works:

• First, take the opposite sign of the constant from the linear polynomial divisor.
• List the coefficients of the polynomial you're dividing.
• Bring down the leading coefficient as it is.
• Multiply the divisor by the value just brought down and add it to the next coefficient.
• Repeat the multiply and add steps until you've processed all coefficients.

This method reduces complexity and organizes calculations neatly, helping achieve an answer without involving variables at every stage. For our problem, it's how we simplify \(s^3 + 10s^2 + 17s + 12\) by \(s+8\).

Remainder Theorem

The Remainder Theorem is a useful tool in understanding polynomial division. It primarily states that if a polynomial \(f(x)\) is divided by \(x-c\), where \(c\) is a constant, the remainder of this division will be \(f(c)\).
Simplified, it predicts the remainder without fully performing the division. In the synthetic division solution, the remainder was found to be 4; confirming it matches what the Remainder Theorem would predict.
When dealing with synthetic division, the usefulness of the Remainder Theorem shines through. Not only does it allow a check of our work, but it also offers a quick way to evaluate polynomials at specific points, providing an assurance that the synthetic division process has been correctly executed.