Question

In Exercises \(1-4,\) find the values of \(A\) and \(B\) that complete the partial fraction decomposition. $$\frac{2 x+16}{x^{2}+x-6}=\frac{A}{x+3}+\frac{B}{x-2}$$

Short answer

The values that complete the partial fraction decomposition are A = 2 and B = 4.

Step-by-step solution

Rewrite the initial equation

Start by rewriting the partial fraction decomposition equation as the following equation where the right side is a single fraction: \[ \frac{2 x+16}{x^2+x-6} = \frac{A(x-2) + B(x+3)}{x^2+x-6} \]

Equating coefficients

By equating above fraction with the left side, we get \[2x + 16 = A(x-2) + B(x + 3)\]. This step helps us break down the problem into a system of equations that can be solved for A and B.

Solve for A and B

To solve for A and B, pick convenient values for x which will allow the equation to be solved. When \(x = 2\), the equation simplifies to \(2*2 + 16 = A(2-2) + B(2+3)\), so B = 4. When \(x = -3\), the equation simplifies to \(2*(-3) + 16 = A(-3-2) + B(-3+3)\), finding A = 2. Thus, we have the solutions A=2 and B=4.

Key concepts

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying mathematical expressions and equations to make them easier to work with. In the context of partial fraction decomposition, algebraic manipulation involves transforming a complex rational expression into a sum of simpler fractions.

For instance, to decompose \[\frac{2 x+16}{x^{2}+x-6}\], we first factor the denominator to identify the linear terms \(x+3\) and \(x-2\). The equation is then expressed as \(\frac{A}{x+3}+\frac{B}{x-2}\), where \(A\) and \(B\) are constants we need to find. The next step involves multiplying each term by the denominator to eliminate the fractions, resulting in \(A(x-2) + B(x + 3)\), which leads us to equate coefficients.

In educational materials, the process of algebraic manipulation should be explained step by step, ensuring that students understand each transformation and simplification. Breaking down each action, such as factoring, distributing, and combining like terms, reinforces algebraic fundamentals crucial for mastery in mathematics.

Equating Coefficients

Equating coefficients is a technique for determining the values of unknown constants in an expression by matching the coefficients of the same powers of the variable. In partial fraction decomposition, once we rewrite the expression as a single fraction with an expanded numerator, we compare the coefficients of corresponding terms from both sides of the equation.

For example, after algebraic manipulation of our initial expression, we have \[2x + 16 = A(x-2) + B(x + 3)\]. Here, we look at the coefficients for \(x\) and the constant terms on both sides of the equation. We create two new equations by collecting terms and equating the coefficients for \(x\) and the constants: one equation from the coefficient of \(x\), and another from the constant term. This is a crucial step, as it transforms the problem into one that can be tackled using methods for solving systems of equations.

In lessons, highlighting the importance of careful organization and comparison of terms is as vital as performing the calculations themselves. This ensures that students grasp the significance of maintaining balance in an equation.

Solving Systems of Equations

Solving systems of equations refers to finding the set of values that satisfy all equations within the system simultaneously. In our partial fraction decomposition, after equating coefficients, we obtain a system of equations that we can solve for \(A\) and \(B\).

In our example, by substituting convenient values for \(x\) into the equation \(2x + 16 = A(x-2) + B(x + 3)\), such as \(x = 2\) and \(x = -3\), we obtain equations that isolate \(A\) and \(B\) respectively. Specifically, \(x = 2\) leads us to \(B = 4\) and \(x = -3\) results in \(A = 2\). These values satisfy the original equation, hence solving the system.

Presenting multiple methods to solve systems, such as substitution, elimination, or matrix operations, and explaining where each is applicable, can fortify a student's problem-solving skills. Furthermore, emphasizing the value of choosing strategic values for \(x\) to simplify calculations can be a helpful problem-solving strategy for students to adopt.