Question

Resolve \(\frac{2 x^{2}+3 x+18}{(x-2)(x+2)^{2}}\) into partial fractions. (1) \(\frac{2}{(x+2)}-\frac{5}{(x-1)^{2}}\) (2) \(\frac{2}{x-2}+\frac{1}{(x+2)}+\frac{3}{(x+2)^{2}}\) (3) \(\frac{2}{x-2}-\frac{5}{(x+2)^{2}}\) (4) \(\frac{2}{x-2}-\frac{1}{(x+2)}+\frac{3}{(x+2)^{2}}\)

Short answer

Answer: $$\frac{2x^2 + 3x + 18}{(x-2)(x+2)^{2}} = \frac{2}{x-2} + \frac{1}{(x+2)^2}$$

Step-by-step solution

Identify the partial fractions form

Based on the denominator, the partial fractions form should resemble the given options, with unknown constants in the numerators: $$\frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$

Clear the denominators

To find the constants, multiply both sides of the equation by the common denominator \(\displaystyle (x-2)(x+2)^{2}\): $$2x^2 + 3x + 18 = A(x+2)^2 + B(x-2)(x+2) + C(x-2)$$

Compare coefficients

Expand the expression and gather like terms on both sides: $$2x^2 + 3x + 18 = Ax^2 + 2Ax + 4A + Bx^2 - 4B + Cx - 2C$$ Now, we should compare coefficients on each side: $$2x^2 + 3x + 18 = (A+B)x^2 + (2A + C)x + (4A - 4B - 2C)$$ This leads to a system of equations: $$A + B = 2$$ $$2A + C = 3$$ $$4A - 4B - 2C = 18$$

Solve the system of equations

We'll use either substitution or elimination methods to solve this system of linear equations to get the values of A, B, and C: Solving the first equation, we get: $$B = 2 - A$$ Substitute this expression for B into the third equation and solve for C in terms of A: $$4A - 4(2 - A) - 2C = 18$$ $$C = 3 - 2A$$ Now substitute B and C expressions in the second equation: $$2A + 3 - 2A = 3$$ After solving, we get the values: $$A = 2$$ $$B = 0$$ $$C = -1$$

Write the solution in partial fractions form

Plug in the values of A, B, and C back into the partial fractions form: $$\frac{2x^2 + 3x + 18}{(x-2)(x+2)^{2}} = \frac{2}{x-2} + \frac{0}{x+2} + \frac{-1}{(x+2)^2}$$ Simplifying the expression, we get the correct answer: Option (2): $$\frac{2x^2 + 3x + 18}{(x-2)(x+2)^{2}} = \frac{2}{x-2} + \frac{1}{(x+2)^2}$$

Key concepts

Algebra

Algebra is the branch of mathematics that deals with symbols and the rules for manipulating those symbols. In the context of partial fractions, algebra is used to break down complex rational expressions into simpler ones. The process involves identifying the structure of the polynomial in the denominator, as seen in our exercise, and setting up corresponding numerators with unknown constants.

By equating the original polynomial to the sum of these simpler fractions, students can create a system of equations based on matching coefficients. This method aids in finding the numerical values for the constants that yield equivalent expressions. The exercise illustrates this application of algebra in decomposing a complex fraction into a sum of simpler, more manageable parts. This simplification is particularly useful for integration and solving differential equations in calculus.

Polynomial Division

Understanding Polynomial Division

Polynomial division is a technique often used preceding or during the process of partial fraction decomposition. Before we separate a complex fraction into partial fractions, it's important to ensure that the numerator's degree is less than the denominator's. If not, polynomial division must be utilized to rewrite the expression so that it fits the criteria for partial fraction decomposition.

The step-by-step solution provided in the exercise does not require polynomial division since the numerator is already of lesser degree than the denominator. However, in the cases where it is needed, long division or synthetic division can be employed to reduce the degree of the numerator, which simplifies the resulting algebraic task at hand.

System of Equations

Solving a System of Equations in Context

A system of equations is a set of simultaneous equations, usually in several unknowns. In the partial fractions context, after clearing denominators and equating coefficients, we end up with a system of linear equations. Each equation corresponds to a specific degree of x, ensuring that the coefficients on both sides of the equation match.

The exercise solution exemplifies this concept by setting up and solving a system of equations to find the unknown constants A, B, and C. These constants are the numerators of the partial fractions that will sum up to the original polynomial fraction. Techniques such as substitution, elimination, or even matrix methods can be employed to solve these equations. Understanding how to manipulate and solve these systems is vital for mastering partial fractions and much of algebra.