Problem 11 - College Algebra Essentials

Question

write the partial fraction decomposition of each rational expression. $$ \frac{3 x+50}{(x-9)(x+2)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of \( \frac{3 x+50}{(x-9)(x+2)} \) is \( \frac{8}{x-9} - \frac{5}{x+2} \).

Step 1: Decompose into Partial Fractions

Write the expression as the sum of partial fractions. We denote the numerator as \(A\) and \(B\). \n\( \frac{3x+50}{(x-9)(x+2)} = \frac{A}{x-9} + \frac{B}{x+2} \)

Step 2: Express the Equation with a Common Denominator

To find the values for the constants \(A\) and \(B\), multiply each fraction by the total denominator and solve for \(A\) and \(B\). \n Do the calculation: \( 3x+50 = A(x+2) + B(x-9) \)

Step 3: Solve for Constants A and B

To solve for \(A\) and \(B\) we can use a system of equations. Replace \(x\) with values that simplify the equation to solve for \(A\) and \(B\). Select x values that will bring each denominator to zero. Let \(x=-2\) and \(x=9\). These are the roots of the denominator. This gives: \n From \(x= -2\) we get \(A = 8\) and from \(x=9\) we get \(B = -5\).

Step 4: Write Down the Final Answer

Replace \(A\) and \(B\) in the original partial fraction decomposition. This will give the decomposition: \n \( \frac{3x+50}{(x-9)(x+2)} = \frac{8}{x-9} - \frac{5}{x+2} \)

Key Concepts

Rational Expressions

Rational expressions are fractions that have polynomials in both their numerator and denominator. They arise frequently in algebra and calculus, and understanding how to manipulate them is crucial for solving a variety of mathematical problems. In essence, a rational expression is similar to any numerical fraction, but instead of numbers, it contains variables.

For example, the rational expression from our exercise \(\frac{3x+50}{(x-9)(x+2)}\) has a polynomial, \(3x+50\), in its numerator and a product of two binomials, \(x-9\) and \(x+2\), in its denominator. One key aspect of working with rational expressions is simplifying them, which can include factoring polynomials, canceling common factors, and finding equivalent expressions. Partial fraction decomposition is one technique to simplify complex rational expressions by breaking them down into simpler, constituent fractions.

Common Denominator

When dealing with fractions, a common denominator is a shared multiple of the denominators. It's necessary for adding, subtracting, or comparing fractions. The same concept applies to rational expressions. To perform operations with rational expressions, or to decompose them into partial fractions, you need to find their common denominator.

In our solved problem, the common denominator is \( (x-9)(x+2) \), which is simply the product of the individual denominators of the partial fractions we seek. Finding a common denominator allows us to combine the fractions into a single expression or, as in the case of partial fraction decomposition, to separate a complex expression into simpler parts. This step is crucial because it lets us solve for the unknown numerators in the next steps, facilitating the breakdown into simpler fractions.

System of Equations

A system of equations consists of two or more equations involving the same set of variables. The solutions to a system of equations are the values of the variables that satisfy all the equations simultaneously. Systems of equations can be solved in various ways, including substitution, elimination, or graphing.

In partial fraction decomposition, after obtaining an equation with a common denominator, we can treat the numerators as a system of equations. By choosing smart values for \(x\), which are often the roots of the denominator or values that simplify the equation, we can solve for the unknowns in the numerators, typically denoted as \(A\), \(B\), and so on.

For instance, in our example, the roots of the denominators, \(x=-2\) and \(x=9\), are chosen to solve for \(A\) and \(B\). They simplify the equations by turning one fraction at a time to zero, allowing us to isolate and solve for each variable separately. This technique effectively reduces the problem of determining several variables down to solving very simple equations that yield the coefficients \(A=8\) and \(B=-5\).