Problem 31 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}+6 x+3}{(x+1)\left(x^{2}+2 x+2\right)}$$

Step-by-Step Solution

Answer

The solution requires calculating the constants A, B, and C from the system of equations, and substituting them back into the general form of decomposition from Step 2. This final equation will be the partial fraction decomposition of the original rational function. Without the actual calculation, exact values of the constants A, B, and C cannot be obtained.

Step 1: Identify the Factors of the Denominator

The factors of the denominator \((x+1)\left(x^{2}+2 x+2\right)\) are \(x+1\) and \(x^{2}+2x+2\). This rational expression will be broken down into simpler fractions whose denominators are these factors.

Step 2: Write Out the General Form of the Decomposition

The general form of the decomposition is \(\frac{5 x^{2}+6 x+3}{(x+1)\left(x^{2}+2 x+2\right)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+2x+2}\). A, B, and C are constants to be determined. The last factor is quadratic, hence it should be written as \(Bx + C\), not just a single constant.

Step 3: Clear the Fractions

Multiply through by the denominator of the left side to clear fractions: \(5x^{2}+6x+3 = A(x^{2}+2x+2) + (Bx+C)(x+1)\).

Step 4: Collect Terms and Match Coefficients

Expand out the right side, collect like terms, and match coefficients on both sides. This will give a system of linear equations in A, B, and C.

Step 5: Solve the Linear System

Solve the system of linear equations for A, B, and C.

Step 6: Substitute Back to the Decomposition

Substitute the values of A, B, and C obtained from step 5 into the general decomposition from step 2. This will give the actual partial fraction decomposition of the original rational function.

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Much like the fractions with numbers you may be more familiar with, these expressions can often be simplified or re-written for various purposes, such as integration, differentiation, or solving equations. The process of partial fraction decomposition is key when dealing with these expressions.

Take the given rational expression \( \frac{5x^{2}+6x+3}{(x+1)(x^{2}+2x+2)} \). The goal of partial fraction decomposition is to break this complex fraction into simpler, more manageable pieces. This is particularly useful when you encounter integrals in calculus or when you try to find inverse Laplace transforms in differential equations.

Factoring Polynomials

Factoring polynomials is a critical skill when working with rational expressions, since the denominator's factors determine the form of the partial fraction decomposition. In our case, the denominator can be factored into \(x+1)\) and \(x^{2}+2x+2\).

While \(x+1\) is linear and can be factored no further, \(x^{2}+2x+2\) is a quadratic polynomial that doesn't factor nicely into real numbers (it has complex roots). However, for partial fraction decomposition, we don't necessarily need to factor it down to linear terms (unless specified). The point here is to ensure each term in the denominator has a corresponding term in the numerator of your partial fractions.

System of Linear Equations

To determine the constants in the partial fraction decomposition, we must solve a system of linear equations that arises from equating coefficients. This is a fundamental concept in algebra that involves finding the values of unknown variables that satisfy multiple equations at once.

In the given solution, after setting up the equations based on matched coefficients, we would typically use methods like substitution, elimination, or matrix operations to solve for \(A\), \(B\), and \(C\). The solution of this system gives us the specific numbers needed to complete our partial fraction decomposition, turning a daunting rational expression into digestible parts for further analysis or computation in various applications.