Question

Resolve \(\frac{5}{(x+2)(x+3)}\) into partial fractions. (1) \(5\left(\frac{1}{x+2}+\frac{1}{x+3}\right)\) (2) \(5\left(\frac{1}{x+2}-\frac{1}{x+3}\right)\) (3) \(10\left(\frac{1}{x+2}-\frac{1}{x+3}\right)\) (4) \(10\left(\frac{1}{x+2}+\frac{1}{x+3}\right)\)

Short answer

Question: Determine the partial fraction decomposition of the given fraction: $$\frac{5}{(x+2)(x+3)}$$ Answer: The partial fraction decomposition of the given fraction is: $$\frac{5}{(x+2)(x+3)} = \frac{1}{x+2} - \frac{1}{x+3}$$

Step-by-step solution

Clear the denominators

Multiply each side of the equation by the common denominator \((x+2)(x+3)\): $$5 = A(x+3) + B(x+2)$$

Solve for the constants A and B

We will use the method of equating coefficients to solve for A and B. First, expand the equation: $$5 = Ax + 3A + Bx + 2B$$ Combine the x terms: $$5 = (A+B)x + (3A + 2B)$$ Since this equation must hold for all values of x, we can equate the coefficients: 1) \(A + B = 0\) (coefficient of x) 2) \(3A + 2B = 5\) (constant term) Solve for A: $$A = -B$$ Substitute A into the second equation: $$3(-B) + 2B = 5$$ Solve for B: $$-B = 1$$ So, B = -1, and hence, A = 1.

Write the partial fraction decomposition of the original expression

Now that we have the constants A and B, we can write the partial fraction decomposition of the original expression: $$\frac{5}{(x+2)(x+3)} = \frac{1}{x+2} - \frac{1}{x+3}$$

Key concepts

Algebraic Fractions

Algebraic fractions are fractions where the numerator or the denominator (or both) are algebraic expressions—expressions that contain variables, numbers, and arithmetic operations. Similar to numerical fractions, the aim is to simplify these complex fractions into simpler forms where possible. This is where partial fractions decomposition comes in handy, especially when integrating or finding limits in calculus.

Partial fractions decomposition involves breaking down a complex algebraic fraction into simpler fractions that are easier to work with. This technique is useful when dealing with rational expressions, where both the numerator and the denominator are polynomials. The example fundamental to our discussion is \( \frac{5}{(x+2)(x+3)} \), where the complex fraction is decomposed into simpler ones, \( \frac{1}{x+2} \) and \( \frac{-1}{x+3} \).

The process usually involves finding constants that satisfy the equality when the original fraction is expressed as the sum of these simpler fractions. This method is very effective for integration, where integrating simple fractions is often more straightforward than integrating the original algebraic fraction.

Polynomial Equations

Polynomial equations are mathematical expressions in which a polynomial is set equal to another polynomial or a constant. A polynomial is constructed of variables and constants using only addition, subtraction, multiplication, and non-negative whole-number exponents. For example, in our exercise, we've worked with the polynomial \( (x+2)(x+3) \), which is the denominator of our algebraic fraction.

When we decompose algebraic fractions, we often seek to find constants for the numerators of the simpler fractions that'll work for every value of x. In other words, we need the decomposed expression and the original polynomial equation to be equivalent. The process involves setting up a system of linear equations to find these constants, as can be seen in our original exercise. The use of polynomial equations is crucial here as it leads to the system of linear equations we must solve to determine the correct constants for the partial fractions.

System of Linear Equations

A system of linear equations consists of two or more linear equations that share a common set of variables. The goal is usually to find the values of these variables that satisfy all the equations simultaneously. The exercise showcases such a system with the equations obtained from equating coefficients: \( A + B = 0 \) and \( 3A + 2B = 5 \).

Solving this type of system is a fundamental part of algebra, and it can be tackled in various ways, such as graphing, substitution, elimination, or by using matrix operations. In the given exercise, we used substitution to solve for the constants A and B since one equation could be easily manipulated to express one variable in terms of the other. This system forms the heart of the partial fractions decomposition process and mastering this concept is crucial for students to work with polynomial equations in integral calculus and other advanced math topics.