Problem 42 - College Algebra and Calculus: An Applied Approach

Question

Write the function in the form \(f(x)=(x-k) q(x)+r\) for the given value of \(k\), and demonstrate that \(f(k)=r\). $$f(x)=x^{3}+3 x^{2}-7 x-6, \quad k=-\sqrt{2}$$

Step-by-Step Solution

Answer

After performing division, the equation can be rewritten as \(f(x)=(x+\sqrt{2})(x^{2} - \sqrt{2} x - 2\sqrt{2}) -4\sqrt{2}-6\). After substituting \(x\) by \(-\sqrt{2}\), we obtain \(f(-\sqrt{2})=-4\sqrt{2}-6\), which verify that \(f(k) = r\).

Step 1 - Perform the Polynomial Division

We are required to represent \(f(x)\) in the form \((x-k) q(x)+r\), for \(k = -\sqrt{2}\). So, we perform the division of \(f(x) = x^{3}+3x^{2}-7x-6\) by \((x-(-\sqrt{2})) = x+\sqrt{2}\). Calculate the division manually or with a synthetic division method. Be careful with arithmetics involving roots.

Step 2 - Obtain the Quotient and Remainder

After performing the division, we obtain the quotient \(q(x)\) and the remainder \(r\).Basically, the division should give a quotient of \(x^{2}-\sqrt{2}x - 2\sqrt{2}\) and a remainder of \(-4\sqrt{2}-6\). Thus, we rewrite \(f(x)\) as \(f(x)=(x+\sqrt{2})(x^{2}-\sqrt{2}x - 2\sqrt{2}) -4\sqrt{2}-6\).

Step 3 - Verify the Remainder Theorem

Replace \(x\) by \(-\sqrt{2}\) in the equation \(f(x) = (x+\sqrt{2})(x^{2}-\sqrt{2}x - 2\sqrt{2}) -4\sqrt{2}-6\). If we obtain equal value then our calculation is correct.

Key Concepts

Remainder Theorem

The Remainder Theorem is a fundamental concept in algebra that connects polynomial division with evaluating polynomials at specific points. When dividing a polynomial \(f(x)\) by \(x-k\), the remainder will be equal to \(f(k)\). This means that after you perform the division, whatever is left over—the remainder—is simply the value of the polynomial evaluated at \(k\). This handy theorem saves time and is precise in verifying your results.

For example, consider dividing \(f(x) = x^3 + 3x^2 - 7x - 6\) by \(x + \sqrt{2}\) where \(k = -\sqrt{2}\). According to the Remainder Theorem, if you plug \(-\sqrt{2}\) into \(f(x)\), you should get the same remainder obtained from your division. In this problem, we found that the remainder is \(-4\sqrt{2} - 6\). By substituting \(-\sqrt{2}\) into the function, you will verify this value, ensuring your division was performed correctly.

Synthetic Division

Synthetic division is a streamlined method of dividing polynomials, which simplifies the process, especially when dealing with linear divisors of the form \(x-k\). Unlike long division, synthetic division is quick and efficient, using only the coefficients of the polynomial and constant \(k\) to perform the calculations directly.

• It reduces the set-up and space required compared to traditional polynomial long division.
• It provides a systematic approach to find both the quotient and the remainder.
• It's especially useful when applying the Remainder Theorem or when you need a quick verification tool.

Despite its simplicity, synthetic division requires careful arithmetic, particularly when the divisor involves irrational numbers, such as \(\sqrt{2}\), as seen in the given exercise. Keeping signs and operations straight is crucial to ensure a correct outcome.

Polynomial Functions

Polynomial functions consist of variables raised to whole number exponents and exhibit various characteristics depending on the degree and coefficients. They are foundational in algebra and calculus, offering a way to model a wide range of real-world phenomena. The function dealt with in the exercise is a cubic polynomial \(f(x) = x^3 + 3x^2 - 7x - 6\), where the leading term is \(x^3\), indicating it is of degree 3.

Polynomials are used to:

• Show trends and patterns when plotted on a graph, often resulting in smooth, continuous curves.
• Allow complex real-world data to be approximated or simplified for analysis.
• Enable further mathematical exploration into calculus topics such as limits, derivatives, and integrals.

In problems like the exercise, understanding polynomial structure is key to applying division techniques effectively and employing tools such as the Remainder Theorem. Polynomials are incredibly versatile, appearing not only in abstract mathematics but also in physics, engineering, economics, and other fields.