Question

In Exercises \(1-4,\) find the values of \(A\) and \(B\) that complete the partial fraction decomposition. $$\frac{16-x}{x^{2}+3 x-10}=\frac{A}{x-2}+\frac{B}{x+5}$$

Short answer

The values of A and B that complete the partial fraction decomposition are A = -4 and B = 3.

Step-by-step solution

Expand the fractions on the right side

Multiply the denominators of the fractions on the right side by each other to equal the left side denominator. This leads to the following equality: \(16-x = A(x + 5) + B(x - 2)\).

Simplify and rearrange terms

Open the brackets in the equation: \(16 - x = Ax + 5A + Bx - 2B\). Then, rearrange the terms: \(16 - x = (A + B)x + 5A - 2B\).

Compare coefficients and solve

Now, compare coefficients on both sides of the equation. For the terms with x, the coefficient on the left side is -1, and on the right side it is A + B. So, A + B = -1. for the constant terms, the coefficient on the left side is 16, and on the right side, it is 5A - 2B. Hence, 16 = 5A - 2B. Now we have two equations and can solve for A and B.

Solve for A and B

From A + B = -1 express A as A = -1 - B and substitute into the second equation: 16 = 5(-1 - B) - 2B, which simplifies to 16 = -5 - 5B - 2B, or 21 = 7B leading to B = 3. Substituting B = 3 into A + B = -1 gives A = -4.

Key concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in mathematics that involves rearranging, simplifying, and working with equations and expressions to find the desired result. It often includes operations like expanding brackets, combining like terms, and factorization. Let’s look at how these techniques are applied in the context of partial fraction decomposition.

The given exercise requires us to use algebraic manipulation to decompose a rational function into simpler fractions that can be more easily integrated or understood. The first step involves expanding the right side of the given equation after equating it to the left side. Here's how it's done: you multiply out the brackets to combine the terms of A and B with the corresponding x and constant terms.

For instance, when given an equation like \(16-x = A(x + 5) + B(x - 2)\), we expand it to get \(16 - x = Ax + 5A + Bx - 2B\). This expands and rearranges the equation into a form where similar terms are grouped together. At this stage, our goal is to construct an equation where the coefficients of like terms on both sides can easily be compared—which takes us to our next key concept, comparing coefficients.

Solving Linear Equations

Solving linear equations is a crucial part of algebra that deals with finding the value of unknown variables. A linear equation is an equation of the first degree, meaning each term is either a constant or the product of a constant and a single variable.

In the partial fraction decomposition problem, once we've rearranged terms using algebraic manipulation, we obtain a linear system of equations. Continuing from our algebraic manipulation, we had two outcomes: \(A + B = -1\) and \(5A - 2B = 16\). These represent two linear equations with two unknowns, A and B. To solve for these variables, we can use methods such as substitution, elimination, or matrix operations.

In the provided solution, the substitution method was used, expressing A in terms of B from the first equation and then substituting it into the second equation. This ordered approach simplifies the problem into one where we can solve for one variable at a time. Once B is determined, its value is then substituted back into the first equation to find A. By systematically applying the rules for solving linear equations, we can deduce the values of A and B that complete the partial fraction decomposition.

Comparing Coefficients

Comparing coefficients is a technique used to find the values of unknowns by equating the coefficients of corresponding terms on both sides of an equation. It is based on the principle that if two polynomials are equal, then their corresponding coefficients for the same powers of x must also be equal.

Returning to our exercise, after algebraic manipulation, we obtained the following: \(16 - x = (A + B)x + 5A - 2B\). To find A and B, we compare the coefficients of the linear terms (in this case, the x terms) and the constant terms on both sides of the equal sign. This gives us two equations: for the x terms, \(A + B = -1\) and for the constants, \(5A - 2B = 16\).

Once we have equated the coefficients, we are left with a system of linear equations to solve. This is where our knowledge of solving linear equations comes into play. By working through these equations systematically, we can solve for the unknown coefficients A and B, which are essential in understanding and completing the partial fraction decomposition.