Problem 57 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. In finding the electric current in a certain circuit, it is necessary to factor the denominator of \(\frac{2 s}{s^{3}+5 s^{2}+4 s+20} \cdot\) Is (a) \((s-2)\) or (b) \((s+5)\) a factor?

Step-by-Step Solution

Answer

(s+5) is a factor.

Step 1: Identify the Problem

We need to determine if \((s-2)\) or \((s+5)\) is a factor of the polynomial in the denominator \(s^{3} + 5s^{2} + 4s + 20\). This will involve using polynomial division or testing using the Remainder Theorem.

Step 2: Apply the Remainder Theorem

The Remainder Theorem states that if \(x-r\) is a factor of a polynomial \(f(x)\), then \(f(r) = 0\). We will evaluate the polynomial at \(s = 2\) and \(s = -5\) to check for factors.

Step 3: Test (s-2) as a Factor

Substitute \(s = 2\) into the polynomial:\[2^3 + 5(2)^2 + 4(2) + 20 = 8 + 20 + 8 + 20 = 56\]Since \(f(2) = 56 eq 0\), (s-2) is not a factor.

Step 4: Test (s+5) as a Factor

Substitute \(s = -5\) into the polynomial:\[(-5)^3 + 5(-5)^2 + 4(-5) + 20 = -125 + 125 - 20 + 20 = 0\]Since \(f(-5) = 0\), (s+5) is a factor.

Key Concepts

Remainder Theorem

The Remainder Theorem is a helpful tool in polynomial algebra, particularly when checking if a linear expression is a factor of a polynomial. It states that for a polynomial \(f(x)\), if you divide it by \(x - r\), the remainder of this division is \(f(r)\). Consequently, if \(f(r) = 0\), then \(x - r\) is a factor of \(f(x)\).

Let's break down how this theorem helped in the exercise. We were tasked with determining whether \((s-2)\) and \((s+5)\) are factors of the polynomial \(s^{3} + 5s^{2} + 4s + 20\). So, by substituting \(r = 2\) and \(r = -5\) into the polynomial, we calculated \(f(2) = 56\) and \(f(-5) = 0\) respectively. Since \(f(-5) = 0\), \((s+5)\) is a factor, while \(f(2) eq 0\), so \((s-2)\) is not a factor.

In essence, the Remainder Theorem provides an efficient way to test for factors, saving time over more laborious methods like synthetic or long division.

Electric Circuit Analysis

Electric circuit analysis often involves understanding how currents flow within various circuit components, including resistors, capacitors, and inductors. When formulating such analyses, polynomials frequently appear, especially when dealing with elements like impedance in AC circuits.

In the given exercise, identifying factors of the polynomial affects the impedance in the circuit. Impedance is a complex concept that varies with frequency and affects how current flows through the circuit. The approach to factor the polynomial, therefore, may reveal key circuit characteristics, like resonance, which occurs when impedance is minimized, allowing maximum current through.

Thus, factoring polynomials in electric circuits is essential to find the optimum functioning conditions and understand how various components react at different frequencies. It leads to effective circuit design and analysis, ensuring reliable and efficient performance.

Polynomial Division

Polynomial division is an algebraic technique used to divide polynomials, similar to the way we divide numbers in arithmetic. This can be accomplished through methods known as long division or synthetic division, each serving to simplify complex polynomial expressions.

In our exercise, if we didn't use the Remainder Theorem, we would rely on polynomial division to determine if \((s-2)\) or \((s+5)\) were factors of \(s^{3} + 5s^{2} + 4s + 20\). Polynomial division helps to decompose a polynomial into simpler components, revealing factors if the division results in a remainder of zero.

For instance, if by dividing \(s^{3} + 5s^{2} + 4s + 20\) by \((s+5)\) you find that there is no remainder, you confirm that \((s+5)\) is indeed a factor, agreeing with our check using the Remainder Theorem. Therefore, polynomial division and the Remainder Theorem are complementary techniques to solve polynomial factorization problems.