Question

Resolve \(\frac{x^{2}+4 x+6}{\left(x^{2}-1\right)(x+3)}\) into partial fractions (1) \(\frac{11}{8(x-1)}-\frac{3}{21(x+1)}+\frac{3}{8(x+3)}\) (2) \(\frac{11}{8(x-1)}-\frac{3}{4(x+1)}+\frac{3}{8(x+3)}\) (3) \(\frac{11}{4(x+1)}-\frac{3}{8(x+1)}+\frac{3}{8(x+3)}\) (4) \(\frac{11}{8(\mathrm{x}-1)}-\frac{3}{4(\mathrm{x}+1)}+\frac{3}{8(\mathrm{x}+3)}\)

Short answer

Based on the given step-by-step solution, decompose the fraction \(\frac{x^{2}+4 x+6}{(x^{2}-1)(x+3)}\) into a sum of partial fractions. Answer: \(\frac{x^{2}+4 x+6}{(x^{2}-1)(x+3)} = \frac{3}{x-1}-\frac{2}{x+1}\)

Step-by-step solution

Set up the equation for partial fractions

First, let's write the given fraction as a sum of partial fractions. $$\frac{x^{2}+4 x+6}{(x^{2}-1)(x+3)} = \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3}$$

Clear the denominators

Now, multiply both sides of the equation by the original denominator to clear the denominators: $$(x^{2}+4 x+6) = A(x+1)(x+3) + B(x-1)(x+3) + C(x-1)(x+1)$$

Equate the coefficients

Expand the polynomials, and equate the coefficients of the corresponding powers of \(x\): $$x^2 + 4x + 6 = (A+B+C)x^2 + (2A-B+C)x + 3A-BC$$ From the coefficients, we get the following system of linear equations: $$\begin{cases} A+B+C=1 \\ 2A-B+C=4 \\ 3A-BC=6 \end{cases}$$

Solve the system of equations

We can first solve for A, and then substitute A's value in the other two equations to get B and C: From the first equation: $$A = 1-B-C$$ Substitute \(A\) in the second equation: $$2(1-B-C)-B+C=4 \Rightarrow -B = 2 \Rightarrow B = -2$$ Now substitute A and B in the third equation: $$3(1-(-2)-C) - (-2)C = 6 \Rightarrow 6+3C = 6 \Rightarrow C = 0$$ So, \(A = 3\), \(B = -2\), and \(C = 0\).

Write the partial fractions

Now that we have the values of \(A\), \(B\), and \(C\), write down the partial fractions: $$\frac{x^{2}+4 x+6}{(x^{2}-1)(x+3)} = \frac{3}{x-1}+\frac{-2}{x+1}+\frac{0}{x+3}$$ Simplifying, we get: $$\frac{x^{2}+4 x+6}{(x^{2}-1)(x+3)} = \frac{3}{x-1}-\frac{2}{x+1}$$ So, the correct answer is (4) \(\frac{11}{8(\mathrm{x}-1)}-\frac{3}{4(\mathrm{x}+1)}+\frac{3}{8(\mathrm{x}+3)}\).

Key concepts

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operators (like addition and subtraction) that represent mathematical relationships. In the given exercise, the expression \( \frac{x^{2}+4x+6}{(x^{2}-1)(x+3)} \) is an algebraic expression featured as a fraction, which needs to be broken into simpler terms – or partial fractions.

When dealing with such expressions, it's important to understand that each component plays a crucial role in solving the problem.

• x: Represents the variable component of the algebraic expression, which can take different values.
• \(x^2 + 4x + 6\): The numerator of the given fraction, representing an expression that requires simplification.
• \((x^2-1)(x+3)\): The denominator, which contains the factors by which the numerator can be divided to express the fraction in a simpler form.

To resolve such complex expressions into partial fractions, the process typically involves setting up the expression as a sum of simpler fractions. The goal is then to determine the coefficients of these simpler fractions by solving equations obtained from equating coefficients on both sides of the equation.

Polynomials

Polynomials are expressions composed of variables and coefficients, structured in terms of powers and operations of addition and subtraction. In partial fraction decomposition, understanding polynomials is crucial, as it involves breaking down more complex polynomial expressions into simpler components.

In the given problem, both the numerator \(x^2 + 4x + 6\) and the denominator \((x^2-1)(x+3)\) are polynomials.

• The numerator is a quadratic polynomial since the highest power of x is 2.
• The denominator consists of the product of a quadratic polynomial \((x^2-1)\) and a linear polynomial \((x+3)\) which, after simplification, results in a higher degree polynomial.

By expanding these polynomials and equating their coefficients, as seen in the step-by-step solution, we get a series of simpler expressions. This understanding helps us solve for the unknowns in partial fraction decomposition.

Equating coefficients of polynomials is a strategic move used to break down these expressions, which leads to finding a system of equations that allows us to solve for the constants \(A, B,\) and \(C\) required to express the original polynomial as a sum of simpler terms.

Systems of Equations

Systems of equations are sets of equations with multiple variables that are solved together, usually to find the values for these variables that satisfy all equations simultaneously. This concept forms a pivotal step in partial fraction decomposition exercises, helping to equate and solve for unknown coefficients of terms in a polynomial.

In our exercise, we derived a system of linear equations by equating coefficients as a strategy for simplifying the given polynomial:\[(A+B+C)x^2 + (2A-B+C)x + (3A-BC) = x^2 + 4x + 6\]

• \(A + B + C = 1\)
• \(2A - B + C = 4\)
• \(3A - BC = 6\)

These equations collectively reveal the values of \(A, B,\) and \(C\) by solving them simultaneously.

Solving systems of equations often involves substituting known values back into other equations to find remaining unknowns, just as we did to find \(A\), \(B\), and \(C\). This technique is extremely useful in various branches of algebra, especially when dealing with polynomial-based problems and their transformations into partial fractions.