Problem 7 - Discrete Mathematics with Applications

Question

Express each quotient as a sum of partial fractions. $$\frac{x^{3}+x^{2}+x+3}{x^{4}+5 x^{2}+6}$$

Step-by-Step Solution

Answer

The given quotient can be expressed as a sum of partial fractions as follows: \[\frac{x^3 + x^2 + x + 3}{(x^2 + 2)(x^2 + 3)} = \frac{x - 2}{x^2 + 2} + \frac{3}{x^2 + 3}\]

Step 1: Factor the denominator

First, we need to factor the denominator: \(x^{4} + 5x^2 + 6\). We notice that this polynomial can be treated as a quadratic polynomial in \(x^2\). With that in mind, it can be factored as: \[x^{4} + 5x^2 + 6 = (x^2 + 2)(x^2 + 3)\] Now that the denominator is factored, we can move on to the next step.

Step 2: Decompose the quotient into partial fractions

Now that we've factored the denominator, we can decompose our rational function into partial fractions, as follows: \[\frac{x^3 + x^2 + x + 3}{(x^2 + 2)(x^2 + 3)} = \frac{Ax + B}{x^2 + 2} + \frac{Cx + D}{x^2 + 3}\]

Step 3: Solve for the constants A, B, C, and D

We need to find the values of the constants \(A, B, C\), and \(D\). To do this, first, clear the denominators by multiplying both sides of the equation by \((x^2 + 2)(x^2 + 3)\). This gives us: \[x^3 + x^2 + x + 3 = (Ax + B)(x^2 + 3) + (Cx + D)(x^2 + 2)\] Now, we'll expand the equation and collect like terms: \[\begin{aligned} x^3 + x^2 + x + 3 = & A \cdot x^3 + (3A + 2C) x^2 + (Ax + B + Cx + D)x + 3B + 2D \end{aligned}\] Next, we'll compare coefficients on both sides of the equation to create a system of linear equations: \[\begin{cases} A = 1 \\ 3A + 2C = 1 \\ A + B + C + D = 1 \\ 3B + 2D = 3 \end{cases}\] Solving this system of equations, we find that \(A=1, B=-2, C=0\), and \(D=3\). Finally, substitute the values of A, B, C, and D into our partial fractions equation: \[\frac{x^3 + x^2 + x + 3}{(x^2 + 2)(x^2 + 3)} = \frac{(1)x + (-2)}{x^2 + 2} + \frac{(0)x + 3}{x^2 + 3}\] Or, simplifying: \[\frac{x^3 + x^2 + x + 3}{(x^2 + 2)(x^2 + 3)} = \frac{x - 2}{x^2 + 2} + \frac{3}{x^2 + 3}\] Thus, we've successfully expressed the given quotient as a sum of partial fractions.

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill when dealing with expressions like the one in this exercise. Essentially, factoring involves rewriting a polynomial as a product of simpler polynomials. This is similar to decomposing numbers into their prime factors.

In our example, the polynomial in question was the denominator: \(x^4 + 5x^2 + 6\). One effective strategy for such expressions is treating them as a quadratic in disguise. Here, \(x^4\) is \((x^2)^2\) and so we can rewrite our expression and factor it as:

• Observe the structure: \(x^4 + 5x^2 + 6\)
• Treat \(x^2\) as \(Y\), so it becomes \(Y^2 + 5Y + 6\)
• Factor like a standard quadratic: \((Y + 2)(Y + 3)\)
• Substitute back: \((x^2 + 2)(x^2 + 3)\)

Factoring provides a simpler path to work with and helps to capitalize on simpler polynomials, making subsequent operations more manageable.

System of Linear Equations

Solving systems of linear equations is a systematic method that aids in finding the unknowns in expressions. In this case, it involves determining the coefficients \(A, B, C\), and \(D\) for our partial fractions. Each step follows logically until all coefficients are identified.

Once the equations stemming from comparing coefficients are set up, we solve them as follows:

• Write down all equations gleaned from matching coefficients.
• In our case, four unknowns and four equations means the problem is well-possible to solve systematically.
• Use methods like substitution or elimination to find each unknown.
• Educated guessing, checking, and trial-and-error can also aid in pinpointing the right values from constraints.

Mastery over this process opens up many algebraic tasks, reinforcing understanding across mathematical problems.

Expansion of Polynomials

The expansion of polynomials is the reverse process of factoring, where you multiply polynomials to simplify expressions or generate original equations from factored forms.

In our exercise, to align terms and solve for unknowns, we expanded:

• Each part of the decomposed fraction was expanded to clear denominators \((Ax + B)(x^2 + 3) + (Cx + D)(x^2 + 2)\).
• After expanding, this became a polynomial in standard form.
• By aligning terms with the polynomial on the other side of the equation, expressing it reveals a map to compare like terms.

Expansion reveals hidden relationships between individual elements, essential for ensuring both sides of an equation represent the same polynomial structure.

With expansion, you can directly confirm operations like addition, multiplication, and distribution, rooting them firmly in reliable algebraic principles.